{
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    "slug": "calm-rocket",
    "projectID": "global",
    "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1",
    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
    "model": {
      "id": "gpt-6-astra",
      "providerID": "openai",
      "variant": "low"
    },
    "version": "1.18.30",
    "summary": {
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    "tokens": {
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      "output": 5005,
      "reasoning": 562,
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    "time": {
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  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
          "created": 1789625269894
        },
        "agent": "nudge-generation",
        "model": {
          "providerID": "openai",
          "modelID": "gpt-6-astra",
          "variant": "low"
        },
        "system": "You create Codeforces learning content for Nudge. Your job is to understand the best approach for the problem and teach it clearly in your own words, using existing solutions to inform your reasoning.\n\nStart with a brief, targeted web search for existing solutions so relevant editorials are in context before substantial independent problem-solving. Use any supplied tutorial links first; otherwise search by contest ID and problem index. Prefer the official tutorial/editorial, and read relevant explanations rather than relying on search snippets. Look at accessible alternatives when useful, especially if the official explanation is missing, unclear, or unnecessarily complicated. Accepted submissions can also help clarify an approach. Respect supplied source lookup statuses and avoid repeatedly retrying blocked sources. If no usable source is accessible, proceed from the supplied statement.\n\nTreat existing solutions as a starting point, not a constraint. If an approach is clear and suitable, build on it without unnecessary re-derivation. If an explanation has gaps, seems incorrect, or uses a harder method than necessary, reason through it and derive a better or simpler approach when worthwhile. Choose a correct approach that is easy to teach and fits the constraints. Verify it against the supplied statement and edge cases before writing the hints, editorial, and C++ solution.\n\nThe editorial must be self-contained: explain the key insight, algorithm, why it works, complexity, and implementation details a learner needs, filling in steps the source skips. Keep research and verification concise; spend the output on teaching the solution, not narrating the research process. Follow the user prompt and output schema exactly.",
        "format": {
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          "schema": {
            "type": "object",
            "properties": {
              "status": {
                "type": "string",
                "enum": [
                  "success",
                  "unsolvable"
                ],
                "description": "Set to 'success' when the supplied statement defines a solvable programming problem. Set to 'unsolvable' only if the statement is fundamentally incomplete, contradictory, or depends on an inaccessible required resource."
              },
              "reason": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 2000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'unsolvable', provide a detailed failure message to show users. Name the concrete statement/resource blocker and any relevant source lookup/access status, such as missing official tutorial, 403/Cloudflare challenge, 404, or unavailable accepted submissions. Otherwise null."
              },
              "hints": {
                "anyOf": [
                  {
                    "type": "array",
                    "minItems": 5,
                    "maxItems": 5,
                    "description": "Exactly 5 progressive hints, each building on the last.",
                    "items": {
                      "type": "object",
                      "properties": {
                        "order": {
                          "type": "integer",
                          "minimum": 1,
                          "maximum": 5,
                          "description": "Hint number, 1 through 5."
                        },
                        "content": {
                          "type": "string",
                          "pattern": "^[^\\u0000]*$",
                          "minLength": 1,
                          "maxLength": 10000,
                          "description": "Markdown hint text."
                        }
                      },
                      "required": [
                        "order",
                        "content"
                      ],
                      "additionalProperties": false
                    }
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', provide exactly 5 progressive hints. Otherwise null."
              },
              "editorial": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 100000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a prose editorial explaining the solution in Nudge's own voice, with no links, citations, or source/research notes. Otherwise null."
              },
              "solution": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 200000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a complete C++ solution. Otherwise null."
              }
            },
            "required": [
              "status",
              "reason",
              "hints",
              "editorial",
              "solution"
            ],
            "additionalProperties": false,
            "title": "problem_response",
            "description": "Submit the generated content for the problem, or report if it is unsolvable."
          },
          "retryCount": 2
        },
        "id": "msg_0adfa7e86001TUn2YqSKz3wMHK",
        "sessionID": "ses_f520583c9ffeiTq6VbX8TnWO8h"
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      "parts": [
        {
          "type": "text",
          "text": "Generate Codeforces learning content: five progressive hints, an original editorial, and AC-quality C++.\n\nFollow the editorial-first research workflow in the system instructions. Use the supplied statement to verify that the source solves this exact problem.\n\nWrite clean Markdown with LaTeX as needed. Use quick and clever humor when appropriate. Tell it like it is (don't sugar-coat responses), and use very casual language. You are fully allowed to swear, just don't overdo it like a sailor (be natural). Deconstruct any false assumptions.\n\nGenerate content for Codeforces problem 2258F: \"Plus Minus Tree\" (rated 2500).\n\nKnown problem metadata:\n- Contest ID: 2258\n- Index: F\n- Name: Plus Minus Tree\n- Rating: 2500 (Codeforces difficulty rating)\n- Tags: data structures, dfs and similar, dp, graphs, greedy, trees\n- Problem URL: https://codeforces.com/contest/2258/problem/F\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n<problem-statement>\n<div class=\"header\"><div class=\"title\">F. Plus Minus Tree</div><div class=\"time-limit\"><div class=\"property-title\">time limit per test</div>3 seconds</div><div class=\"memory-limit\"><div class=\"property-title\">memory limit per test</div>256 megabytes</div><div class=\"input-file input-standard\"><div class=\"property-title\">input</div>standard input</div><div class=\"output-file output-standard\"><div class=\"property-title\">output</div>standard output</div></div><div><p>     </p><p>You are given a tree $$$T$$$ with $$$n$$$ vertices, rooted$$$^{\\text{∗}}$$$ at vertex $$$1$$$. Each vertex $$$v$$$ is assigned an initial weight $$$ a_v \\in \\{ -1, 0, +1 \\} $$$.</p><p>For each vertex with weight $$$0$$$, you have to assign a new weight, either $$$-1$$$ or $$$+1$$$. Let $$$x_v$$$ be the final weight of vertex $$$v$$$.</p><p>For vertex $$$v$$$, let $$$T_v$$$ be the set of nodes in the subtree$$$^{\\text{†}}$$$ of vertex $$$v$$$. The cost of vertex $$$v$$$ is defined as $$$$$$ S_v = \\left| \\sum_{u \\in T_v}{x_u} \\right|.$$$$$$</p><p>The total cost of the tree is defined as the sum of costs over all vertices. Find the minimum possible total cost of the tree after assigning the weights of all vertices with zero initial weight.</p><div class=\"statement-footnote\"><p>$$$^{\\text{∗}}$$$A tree is a connected graph without cycles. A rooted tree is a tree where one vertex is special and called the root.</p><p>$$$^{\\text{†}}$$$A subtree of vertex $$$v$$$ is the subgraph consisting of $$$v$$$, all its descendants, and all the edges between them. A descendant of vertex $$$v$$$ is any vertex $$$u$$$ for which $$$v$$$ is an ancestor. No vertex is its own descendant. An ancestor of vertex $$$v$$$ is any vertex on the simple path from $$$v$$$ to the root, including the root, but not including $$$v$$$. The root has no ancestors.</p></div></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 10^4$$$). The description of the test cases follows.</p><p>The first line of each test case contains $$$n$$$ ($$$2 \\le n \\le 2 \\cdot 10^5$$$), denoting the number of vertices in the tree. </p><p>The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \\ldots, a_n$$$ ($$$-1 \\le a_i \\le 1$$$). </p><p>Each of the next $$$n-1$$$ lines contains two integers $$$u$$$ and $$$v$$$ ($$$1 \\le u, v \\le n$$$), denoting the nodes connected with an edge of the tree. It is guaranteed that the given edges form a tree.</p><p>It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \\cdot 10^5$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For each test case, print the minimum total cost of the tree.</p></div><div class=\"sample-tests\"><div class=\"section-title\">Example</div><div class=\"sample-test\"><div class=\"input\"><div class=\"title\">Input</div><pre><div class=\"test-example-line test-example-line-even test-example-line-0\">2</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">5</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">0 1 -1 0 1</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1 2</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">1 3</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">2 4</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">2 5</div><div class=\"test-example-line test-example-line-even test-example-line-2\">8</div><div class=\"test-example-line test-example-line-even test-example-line-2\">1 -1 1 -1 1 -1 1 -1</div><div class=\"test-example-line test-example-line-even test-example-line-2\">1 2</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 3</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 4</div><div class=\"test-example-line test-example-line-even test-example-line-2\">3 5</div><div class=\"test-example-line test-example-line-even test-example-line-2\">4 6</div><div class=\"test-example-line test-example-line-even test-example-line-2\">6 7</div><div class=\"test-example-line test-example-line-even test-example-line-2\">2 8</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre><div class=\"test-example-line test-example-line-odd test-example-line-1\">5</div><div class=\"test-example-line test-example-line-even test-example-line-2\">7</div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>The best assignments in the first test case are $$$[+1, +1, -1, -1, +1]$$$ and $$$[-1, +1, -1, -1, +1]$$$. They have a cost of $$$5$$$.</p><p>Another assignment, $$$[+1, +1, -1, +1, +1]$$$ has a cost of $$$9$$$, and $$$[-1, +1, -1, +1, +1]$$$ has a cost of $$$7$$$.</p><center> <table class=\"tex-tabular\"><tbody><tr><td class=\"tex-tabular-text-align-center tex-tabular-border-right\"><img class=\"tex-graphics\" src=\"https://espresso.codeforces.com/e6d79fb7b6e201ed3e9f3c159988cbb6734c6976.png\" style=\"max-width: 100.0%;max-height: 100.0%;\"></td><td class=\"tex-tabular-border-left tex-tabular-text-align-center\"><img class=\"tex-graphics\" src=\"https://espresso.codeforces.com/f4caa2b2809df78444a8b72a7df27fd522569266.png\" style=\"max-width: 100.0%;max-height: 100.0%;\"></td><td></td></tr><tr><td class=\"tex-tabular-text-align-center tex-tabular-border-right\"><span class=\"tex-font-size-small\">The tree of the first testcase</span></td><td class=\"tex-tabular-border-left tex-tabular-text-align-center\"><span class=\"tex-font-size-small\">The tree of the second testcase</span></td><td></td></tr></tbody></table> </center></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/156316 loaded (200 OK; title: \"Codeforces Round 1118 (Div. 2) Editorial - Codeforces\"; mentions 2258F)\n</source-lookup-status>\n\nGenerate:\n1. Five progressive hints, from a gentle nudge to the key insight.\n2. A self-contained editorial that teaches the solution clearly, including the key insight, algorithm, correctness argument, complexity, and necessary implementation details. Explain non-obvious steps without padding or a discovery-process transcript.\n3. A complete C++26 solution that gets AC on Codeforces.\n\nFormatting & Style Rules:\n- Hints and editorial: valid Markdown with LaTeX as needed (e.g., $dp[i]$, $$\\sum_{i=1}^{n} a_i$$).\n- Hints and editorial must read like Nudge's own explanation. No research notes, source notes, citations, Markdown links, URLs, or references to editorials/submissions/posts.\n- Solution: raw C++ only, no Markdown fences.\n- Keep C++ short, clean, standard, and single-file. Comments are fine when they clarify the idea.\n\nOutput strictness:\n- Return JSON matching the provided schema exactly.\n- Never include NUL (U+0000) characters in any field. Use literal $ or $$ for math delimiters, never control characters.\n- If the problem is solvable, return `status: \"success\"`, `reason: null`, and fill in `hints`, `editorial`, and `solution`.\n- Return `status: \"unsolvable\"`, a short `reason`, and set `hints`, `editorial`, and `solution` to null only when the supplied statement is fundamentally incomplete, contradictory, or dependent on an inaccessible resource needed to define the task. Missing research, high rating, uncertainty, or \"can't guarantee AC\" is not a reason to return unsolvable.\n- If you return `status: \"unsolvable\"`, the `reason` is shown directly to users. Be concrete: say what exact statement/resource blocker stopped you, and include relevant source lookup status if it mattered (for example: official tutorial missing, Codeforces returned 403/Cloudflare challenge, tutorial page 404, accepted submissions unavailable).\n- Each hint must be an object with `order` (1 through 5) and `content` (the Markdown hint text), not a plain string. The `content` must contain just the hint text, with no \"Hint 1:\" or subtitles. The UI adds those automatically.\n- Do not start the editorial with an \"# Editorial\" heading. The UI already adds that section. Feel free to add other headers as needed though.\n- Use styling in the editorial if needed! Just that first header is no good but any subheaders or bolding or whatever is good!\n\nFor the C++ solution, you MUST use this template and work around it:\n\n```cpp\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n}\n```",
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",
          "mime": "image/png",
          "filename": "image-1.png",
          "id": "prt_0adfa7e8b003y9afCSr0nxLRoz",
          "sessionID": "ses_f520583c9ffeiTq6VbX8TnWO8h",
          "messageID": "msg_0adfa7e86001TUn2YqSKz3wMHK"
        },
        {
          "type": "text",
          "synthetic": true,
          "text": "Called the Read tool with the following input: {\"filePath\":\"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1/assets/image-2.png\"}",
          "id": "prt_0adfa7e8b004erDrDkxTrwH85D",
          "sessionID": "ses_f520583c9ffeiTq6VbX8TnWO8h",
          "messageID": "msg_0adfa7e86001TUn2YqSKz3wMHK"
        },
        {
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",
          "mime": "image/png",
          "filename": "image-2.png",
          "id": "prt_0adfa7e8b005Jmy2A8gRLwWZlE",
          "sessionID": "ses_f520583c9ffeiTq6VbX8TnWO8h",
          "messageID": "msg_0adfa7e86001TUn2YqSKz3wMHK"
        }
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    },
    {
      "info": {
        "parentID": "msg_0adfa7e86001TUn2YqSKz3wMHK",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "low",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1",
          "root": "/"
        },
        "cost": 0,
        "tokens": {
          "total": 11245,
          "input": 11213,
          "output": 32,
          "reasoning": 0,
          "cache": {
            "write": 0,
            "read": 0
          }
        },
        "modelID": "gpt-6-astra",
        "providerID": "openai",
        "time": {
          "created": 1789625270049,
          "completed": 1789625277608
        },
        "finish": "tool-calls",
        "id": "msg_0adfa7f21001vmsFs25PwBMoom",
        "sessionID": "ses_f520583c9ffeiTq6VbX8TnWO8h"
      },
      "parts": [
        {
          "type": "step-start",
          "id": "prt_0adfa8d4e001hxRWnnJyFwl0OF",
          "sessionID": "ses_f520583c9ffeiTq6VbX8TnWO8h",
          "messageID": "msg_0adfa7f21001vmsFs25PwBMoom"
        },
        {
          "type": "tool",
          "tool": "webfetch",
          "callID": "call_NP8NDPpBKRMmAWFdFgbidiI0",
          "state": {
            "status": "completed",
            "input": {
              "url": "https://codeforces.com/blog/entry/156316",
              "format": "markdown"
            },
            "output": "---\nimage: https://codeforces.com/codeforces.org/s/85295/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1118 (Div. 2) Editorial - Codeforces\ndescription: Codeforces. Programming competitions and contests, programming community\n---\n\n<!DOCTYPE HTML PUBLIC \"-//W3C//DTD HTML 4.01//EN\"> \n\n  \n[![Codeforces](//codeforces.com/codeforces.org/s/85295/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/85295/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/156316?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/85295/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/156316?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F156316)| [Register](https://codeforces.com/register) \n\n  \n* [Home](https://codeforces.com/)\n* [Top](https://codeforces.com/top)\n* [Catalog](https://codeforces.com/catalog)\n* [Contests](https://codeforces.com/contests)\n* [Gym](https://codeforces.com/gyms)\n* [Problemset](https://codeforces.com/problemset)\n* [Groups](https://codeforces.com/groups)\n* 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-------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------- | -------------------------------------------- |\n\n→ Top contributors \n\n| #  | User                                                                                     | Contrib. |\n| -- | ---------------------------------------------------------------------------------------- | -------- |\n| 1  | [Qingyu](https://codeforces.com/profile/Qingyu \"Legendary Grandmaster Qingyu\")           | 157      |\n| 2  | [maspy](https://codeforces.com/profile/maspy \"Legendary Grandmaster maspy\")              | 150      |\n| 2  | [nik\\_exists](https://codeforces.com/profile/nik%5Fexists \"Expert nik_exists\")           | 150      |\n| 4  | [Um\\_nik](https://codeforces.com/profile/Um%5Fnik \"Legendary Grandmaster Um_nik\")        | 144      |\n| 5  | [Errichto](https://codeforces.com/profile/Errichto \"International Grandmaster Errichto\") | 139      |\n| 6  | [adamant](https://codeforces.com/profile/adamant \"Master adamant\")                       | 136      |\n| 7  | [AmShZ](https://codeforces.com/profile/AmShZ \"Grandmaster AmShZ\")                        | 134      |\n| 8  | [maroonrk](https://codeforces.com/profile/maroonrk \"Legendary Grandmaster maroonrk\")     | 133      |\n| 9  | [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") | 132      |\n| 10 | [qwexd](https://codeforces.com/profile/qwexd \"Master qwexd\")                             | 129      |\n\n|  | [View all →](https://codeforces.com/top-contributed) |\n|  | ---------------------------------------------------- |\n\n→ Find user \n\nHandle: \n\n→ Recent actions \n\n* [ko\\_osaga](https://codeforces.com/profile/ko%5Fosaga \"Grandmaster ko_osaga\") → [Response to \"Is IOI Becoming Too Heuristic?\"](https://codeforces.com/blog/entry/156329) ![New 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[Submissions](https://codeforces.com/submissions/ItsNotMeItsYou)\n* [Groups](https://codeforces.com/groups/with/ItsNotMeItsYou)\n* [Contests](https://codeforces.com/contests/with/ItsNotMeItsYou)\n* [Problemsetting](https://codeforces.com/contests/writer/ItsNotMeItsYou)\n\n### [ItsNotMeItsYou's blog](https://codeforces.com/blog/ItsNotMeItsYou)\n\n[ Codeforces Round 1118 (Div. 2) Editorial ](https://codeforces.com/blog/entry/156316) \n\nBy [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\"), 3 weeks ago, ![In English](//codeforces.com/codeforces.org/s/85295/images/flags/24/gb.png \"In English\") \n\nSorry for underestimating the difficulties of the problems, especially B2 and D. We tried to serve as many cool problems as we could. And apparently, this led to some difficult ones.\n\nThank you for participating in the round. We all hope you enjoyed all of our problems. \n\n[2258A - Odd Eraser](https://codeforces.com/contest/2258/problem/A \"Codeforces Round 1118 (Div. 2)\") Idea: [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\"), [mychecksdead](https://codeforces.com/profile/mychecksdead \"Master mychecksdead\")\n\n**Hint1**\n\nThink of the case $$$n=3$$$. Can you remove the first or the last number in any way?\n\n**Solution**\n\nWe cannot obtain a number larger than $$$\\\\gcd(a\\_1,a\\_n)$$$ because the first and last numbers in the array can't be removed. We can remove all other elements by choosing $$$\\[1, i, n\\]$$$, so the answer is $$$\\\\gcd(a\\_1, a\\_n)$$$.\n\n**Code**\n\n[388911115](https://codeforces.com/contest/2258/submission/388911115 \"Submission 388911115 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2258B1 - Carrot Chopdown (Easy Version)](https://codeforces.com/contest/2258/problem/B1 \"Codeforces Round 1118 (Div. 2)\") Idea: [mychecksdead](https://codeforces.com/profile/mychecksdead \"Master mychecksdead\")\n\n**Hint1**\n\nWhat is the answer for a fixed $$$x$$$?\n\n**Solution**\n\nIterating over the values $$$x=1...m$$$, we can find the answer for each possible $$$x$$$.\n\nFor any $$$x$$$, the answer is the number of carrots of length greater than or equal to $$$x$$$, plus the number of carrots of length $$$2x$$$.\n\n**Code**\n\n[388911137](https://codeforces.com/contest/2258/submission/388911137 \"Submission 388911137 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2258B2 - Carrot Chopdown (Hard Version)](https://codeforces.com/contest/2258/problem/B2 \"Codeforces Round 1118 (Div. 2)\") Idea: [mychecksdead](https://codeforces.com/profile/mychecksdead \"Master mychecksdead\") (This problem was originally designed with only $$$k=2$$$; thanks to [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\") for expanding on the idea.)\n\n**Hint1**\n\nConsider solving the problem for a fixed $$$x$$$.\n\n**Hint2**\n\nConsider solving the problem for $$$k=2$$$.\n\n**Hint3**\n\nWhat happens if $$$k$$$ is bigger than $$$18$$$? \n\n**Hint4**\n\nWhat are the optimal $$$x$$$ values for cutting?\n\n**Solution**\n\nFor target value $$$x$$$, it is optimal to use cuts in the following order: $$$x \\\\cdot 2^{k-1}, x \\\\cdot 2^{k-2}, \\\\ldots, x \\\\cdot 1$$$.\n\nFor each $$$i$$$, $$$a\\_i=2^k \\\\cdot x$$$ adds $$$2^k$$$ to the answer. Otherwise, the contribution is $$$\\\\min(\\\\frac{a\\_i}{x}, 2^k-1)$$$. \n\nThe optimal $$$x$$$ is less than or equal to $$$\\\\frac{m}{2^k}$$$ (if less than $$$1$$$, it equals $$$1$$$, and the answer is basically the sum of all carrot lengths). Moreover, no $$$i$$$ can contribute more than $$$2^k$$$ to any $$$x$$$. Using prefix sum, determine the number of values contributing $$$y$$$ to the answer for each possible $$$x$$$. This solution runs in $$$\\\\mathcal{O}(n + m \\\\log m)$$$.\n\n**Code**\n\n[388911156](https://codeforces.com/contest/2258/submission/388911156 \"Submission 388911156 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2258C - Far Cities](https://codeforces.com/contest/2258/problem/C \"Codeforces Round 1118 (Div. 2)\") Idea: [Seferoglu](https://codeforces.com/profile/Seferoglu \"Expert Seferoglu\") (This problem was originally designed simply to find the furthest node from the root; thanks to [cadmiumky](https://codeforces.com/profile/cadmiumky \"Grandmaster cadmiumky\") for expanding on the idea. Thanks to [robert9524](https://codeforces.com/profile/robert9524 \"Candidate Master robert9524\") for providing stronger tests on cutting random solutions without an adaptive grader.)\n\n**Hint1**\n\nHow to find a diameter?\n\n**Hint2**\n\nConsider the simplest algorithm.\n\n**Solution**\n\nFor each node, first determine whether it is closer to the first vertex than the longest distance seen so far. If not, just skip the node. This uses $$$n-1$$$ queries. Otherwise, while the query returns true, increase the queried distance by one. This can happen up to $$$n-1$$$ times. We can find one end of the diameter by running $$$2n - 2$$$ queries in total. Then repeat the process to find the other end of the diameter. This yields a total of $$$4n-4$$$ queries. But, as previously written, distance can increase at most $$$n-1$$$ times. So this yields a total of $$$3n-3$$$ queries.\n\n**Code**\n\n[388911205](https://codeforces.com/contest/2258/submission/388911205 \"Submission 388911205 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2258D - Magic Tiles](https://codeforces.com/contest/2258/problem/D \"Codeforces Round 1118 (Div. 2)\") Idea: [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\")\n\n**Hint1**\n\nWhat happens when a range includes another?\n\n**Hint2**\n\nA range that is included by another range is useless. What happens when we remove all such useless ranges?\n\n**Hint3**\n\nWe only need to solve the problem for chains of ranges. How?\n\n**Solution**\n\nIn this editorial we will use BigIntegers, which are simple vectors, and addition and comparison is $$$\\\\mathcal{O}(n)$$$. It can be proven that if one range includes another, we can remove the other. Also, it can be proven that only start and endpoints of the intervals are important, and let them be $$$pts$$$. Well-known $$$\\\\mathcal{O}(n^3)$$$ dp exists; $$$dp\\_i$$$ means the maximum score if we fill all of the rows through $$$pts\\_i$$$; $$$dp\\_i=\\\\max dp\\_j + 100^{100^{i-j}}$$$ if there is an interval that covers interval $$$\\[j, i)$$$. Because there are only $$$2$$$ columns, it can be proven that this solution works in $$$\\\\mathcal{O}(n^2)$$$. From another perspective, you can imagine you have chains. \n\n**Code**\n\n[388911243](https://codeforces.com/contest/2258/submission/388911243 \"Submission 388911243 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n**Challenge**\n\nCan you solve for bigger constraints, such as $$$n,m \\\\le 2 \\\\cdot 10^5$$$? Thanks to [dinohaur](https://codeforces.com/profile/dinohaur \"Master dinohaur\") for the code: [388911329](https://codeforces.com/contest/2258/submission/388911329 \"Submission 388911329 by ItsNotMeItsYou\") \n\n**Answer**\n\n With using faster BigIntegers, you can solve in $$$\\\\mathcal{O}(n \\\\log n)$$$. For more info: <https://codeforces.com/problemset/problem/464/E>\n\n[2258E - DivMEX](https://codeforces.com/contest/2258/problem/E \"Codeforces Round 1118 (Div. 2)\") Idea: [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\"), [carcinisation](https://codeforces.com/profile/carcinisation \"International Master carcinisation\") (This problem was actually designed differently. Thanks to [anpaio](https://codeforces.com/profile/anpaio \"International Grandmaster anpaio\") for this version.)\n\n**Hint1**\n\nWhat kind of integers can exist as an $$$f(l,r)=x$$$ value?\n\n**Hint2**\n\nHow to check a value $$$x$$$ exists as a $$$f(l,r)$$$?\n\n**Solution**\n\nIt can be seen that each value of $$$f$$$ must satisfy $$$f(l,r) \\\\leq \\\\text{smallest prime power bigger than n}$$$; for each $$$x$$$ that satisfies this constraint, we will try to check whether this value can be obtained or not. \n\nA range that satisfies $$$f(l,r)=x$$$ can't contain a multiple of $$$x$$$; otherwise, the set of divisors would have $$$x$$$ in it. When considering multiples of $$$x$$$ as obstacles, we realize that it is optimal to greedily select between two consecutive obstacles. Hence, at each obstacle, we check whether every number $$$y$$$ that satisfies $$$1 \\\\leq y \\\\le x$$$ appears as a factor of any integer between this and the previous obstacle. This solution works in $$$\\\\mathcal{O}(( \\\\sum\\_{i=1}^n |S\\_i|) \\\\cdot n$$$. Where $$$S\\_x$$$ represents the prime powers of $$$x$$$. \n\nTo optimize this solution, we iterate over all $$$a\\_i$$$ and check every factor of $$$a\\_i$$$, considering the index $$$i$$$ as an obstacle. In order to verify that every element smaller than $$$x$$$ is present between this and the previous obstacle, use a minimum segment tree. This solution works in $$$\\\\mathcal{O}(( \\\\sum\\_{i=1}^n |S\\_i|) \\\\log n) \\\\le \\\\mathcal{O}(n \\\\log^2 n)$$$.\n\n**Code (anpaio)**\n\n[388911266](https://codeforces.com/contest/2258/submission/388911266 \"Submission 388911266 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n[2258F - Plus Minus Tree](https://codeforces.com/contest/2258/problem/F \"Codeforces Round 1118 (Div. 2)\") Idea: [ItsNotMeItsYou](https://codeforces.com/profile/ItsNotMeItsYou \"Master ItsNotMeItsYou\")\n\n**Solution**\n\nConsider $$$\\\\mathcal{O}(n^2)$$$ $$$dp\\[v\\]\\[\\\\text{#1's count in subtree of v}\\]$$$ using the merging subtrees technique. The dp is convex because we are adding only two convex functions, resulting in a convex function. Then, using small to large, you can achieve $$$\\\\mathcal{O}(n \\\\log^2 n)$$$ complexity.\n\nTo elaborate: \n\nFor $$$\\\\mathcal{O}(n^2)$$$, you can keep dp as pairs for simplicity: $$$(\\\\text{score of v}, \\\\text{minimum score})$$$. \n\nAs can be seen in the vector, scores are increasing by $$$2$$$. So you can keep dp with the vector's first element and then with the differences. When you merge two children's subtrees, you are basically doing a Minkowski sum, which is just updating the first element as $$$\\\\text{(lhs\\[0\\].first + rhs\\[0\\].first, lhs\\[0\\].second + rhs\\[0\\].second})$$$, and merging the differences. \n\nThen, for each point added to $$$|x|$$$, you must add $$$+2$$$ or $$$-2$$$ ($$$2$$$ due to parity) for all differences split by $$$0$$$ (be careful not to add any difference to $$$\\[-1, +1\\]$$$). \n\nYou can maintain two sets for each vertex, resulting in $$$\\\\mathcal{O}(n \\\\log^2 n)$$$ complexity from small to large. \n\n**Code**\n\n[388911297](https://codeforces.com/contest/2258/submission/388911297 \"Submission 388911297 by ItsNotMeItsYou\")\n\n**Alternative Solution**\n\nFirst, identify all nodes in the tree with an initial value of zero, set them to one, and store them in a candidate list. Then, until the list is completely empty, select the most advantageous node at each step. For each node, we trace the path up to the root, counting how many nodes along that path have a subtree sum greater than one ($$$\\\\text{pos}$$$) and how many have a subtree sum less than one ($$$\\\\text{neg}$$$), thereby calculating a score based on the difference $$$\\\\text{pos} - \\\\text{neg}$$$.\n\nAfter finding the node with the highest score among the list, select it, change its value to minus one, and decrease the subtree sums of all ancestors along the path from the chosen node to the root by two. We repeat this process until all special nodes are processed and the list is empty. And you can fasten this algorithm by HLD. \n\n$$$\\\\mathcal{O}(n^3)$$$ code by [anpaio](https://codeforces.com/profile/anpaio \"International Grandmaster anpaio\"): [388911953](https://codeforces.com/contest/2258/submission/388911953 \"Submission 388911953 by ItsNotMeItsYou\") $$$\\\\mathcal{O}(n \\\\log^2 n)$$$ code by [anpaio](https://codeforces.com/profile/anpaio \"International Grandmaster anpaio\"): [388911806](https://codeforces.com/contest/2258/submission/388911806 \"Submission 388911806 by ItsNotMeItsYou\")\n\n**Rate the problem**\n\nAmazing [ ](https://codeforces.com/data/like?action=like \"I like this\") Good [ ](https://codeforces.com/data/like?action=like \"I like this\") Ok [ ](https://codeforces.com/data/like?action=like \"I like this\") Bad [ ](https://codeforces.com/data/like?action=like \"I like this\") Terrible [ ](https://codeforces.com/data/like?action=like \"I like this\") \n\n![](//codeforces.com/codeforces.org/s/85295/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1118 (Div. 2)](https://codeforces.com/contest/2258) \n\n* [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/voteup-gray.png \"Vote: I like it\")](#)\n* +44\n* [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/votedown-gray.png \"Vote: I do not like it\")](#)\n\n* [![Author](//codeforces.com/codeforces.org/s/85295/images/blog/user_16x16.png \"Author\")](https://codeforces.com/profile/ItsNotMeItsYou) [ItsNotMeItsYou ](https://codeforces.com/profile/ItsNotMeItsYou)\n* ![Publication date](//codeforces.com/codeforces.org/s/85295/images/blog/date_16x16.png \"Publication date\") 3 weeks ago\n* [![Comments](//codeforces.com/codeforces.org/s/85295/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/156316#comments) [76 ](https://codeforces.com/blog/entry/156316#comments)\n\n  \n![Comments](//codeforces.com/codeforces.org/s/85295/images/icons/comments-48x48.png \"Comments\") Comments (76) \n\n[Write comment?](#) \n\n| » [MaxBlazeIceInk](https://codeforces.com/profile/MaxBlazeIceInk \"Candidate Master MaxBlazeIceInk\") | 3 weeks ago, [show](#) # |\n| --------------------------------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/2334734/avatar/fb176e49ae034a0a.jpg) ](https://codeforces.com/profile/MaxBlazeIceInk) [MaxBlazeIceInk](https://codeforces.com/profile/MaxBlazeIceInk \"Candidate Master MaxBlazeIceInk\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389109 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +34 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) On clist.by we can check the hardest div2D since 2026:wonderful 2500 difficulty! [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [rajakrishna](https://codeforces.com/profile/rajakrishna \"Pupil rajakrishna\") | 3 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/5072937/avatar/d8cce84f7422827b.jpg) ](https://codeforces.com/profile/rajakrishna) [rajakrishna](https://codeforces.com/profile/rajakrishna \"Pupil rajakrishna\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389110 \"Link to comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) \\-17 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) . [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [Daniel-Alp](https://codeforces.com/profile/Daniel-Alp \"Pupil Daniel-Alp\") | 3 weeks ago, [show (+1)](#) # |\n| ---------------------------------------------------------------------------- | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/Daniel-Alp) [Daniel-Alp](https://codeforces.com/profile/Daniel-Alp \"Pupil Daniel-Alp\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389112 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I enjoyed the contest! (despite being one of the goobers who killed all their time on B2 :P) [→](#) [Reply](#) |\n| -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [lmjsjg](https://codeforces.com/profile/lmjsjg \"Specialist lmjsjg\") | 3 weeks ago, [show](#) # |\n| ----------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/2917220/avatar/dee6d6ea3380cb56.jpg) ](https://codeforces.com/profile/lmjsjg) [lmjsjg](https://codeforces.com/profile/lmjsjg \"Specialist lmjsjg\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389207 \"Link to comment\") [^](#comment-1389112 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) me too... [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n* | » » [superl61](https://codeforces.com/profile/superl61 \"Newbie superl61\") | 2 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/superl61) [superl61](https://codeforces.com/profile/superl61 \"Newbie superl61\") | 2 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389741 \"Link to comment\") [^](#comment-1389112 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) so do I [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [show (+1)](#) # |\n| -------------------------------------------------------------------- | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/3726532/avatar/64912144a3b5065f.jpg) ](https://codeforces.com/profile/allcaps) [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389117 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +41 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) editorial: \"Thanks to robert9524 for providing stronger tests on cutting random solutions without an adaptive grader\"me: AC'd with a randomised solution and feels like the 0.01% of bacteria that don't get killed by soap [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [Hyder1102](https://codeforces.com/profile/Hyder1102 \"Expert Hyder1102\") | 3 weeks ago, [show (+1)](#) # |\n| ---------------------------------------------------------------------------- | ----------------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/4156580/avatar/3c7005d8f3eefc09.jpg) ](https://codeforces.com/profile/Hyder1102) [Hyder1102](https://codeforces.com/profile/Hyder1102 \"Expert Hyder1102\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389285 \"Link to comment\") [^](#comment-1389117 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +4 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) what was your randomized appraoch? [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n  * | » » » [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [show (+1)](#) # |\n| ------------------------------------------------------------------------ | ----------------------------- |\n\n| » » » [ ![](https://codeforces.com/userpic.codeforces.org/3726532/avatar/64912144a3b5065f.jpg) ](https://codeforces.com/profile/allcaps) [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389297 \"Link to comment\") [^](#comment-1389285 \"Parent comment\") \\| [←](#)Rev. 4 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +9 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) basically because so many interactive problems have binary search, my brain went straight to binary searching for the exact distance when you find a new farthest node, instead of realising that you can just increment it by one and that will happen at most $$$n - 1$$$ times.here's what i ended up doing: same general idea of the two DFS passes in each pass, keep track of furthest node $$$u$$$ and its distance $$$d$$$ maintain a variable $$$h$$$ as the maximum length a path in this graph can have (initially $$$n - 1$$$) when considering a new node, first check if it beats the distance by at least 1\\. if not, stop and go to the next node. otherwise, binary search between $$$\\[d + 1, h\\]$$$ to find distance of this node. then update $$$h$$$ to be $$$MIN(h, \\\\text{new distance + number of nodes left})$$$. the randomised part of the solution is that instead of starting at vertex $$$1$$$, and then querying $$$2, 3, 4, ... n$$$ in order, I shuffle the vertices so that I should only need to do $$$O(log(n))$$$ binary searches, so on average, I should have $$$2 n - 2 + O(log^2(n))$$$ queries. The worst case is still just $$$n log(n)$$$. intuitively, the worst case for such a solution should just be a path, so i wrote a program locally to simulate the process on a bunch of randomised paths, and it never exceeded $$$3n$$$ queries there. submission: [388829880](https://codeforces.com/contest/2258/submission/388829880 \"Submission 388829880 by allcaps\") [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n    * | » » » » [sergioRG](https://codeforces.com/profile/sergioRG \"Expert sergioRG\") | 3 weeks ago, [show (+1)](#) # |\n| ----------------------------------------------------------------------------- | ----------------------------- |\n\n| » » » » [ ![](https://codeforces.com/userpic.codeforces.org/82000/avatar/6a2b34040eea0d59.jpg) ](https://codeforces.com/profile/sergioRG) [sergioRG](https://codeforces.com/profile/sergioRG \"Expert sergioRG\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389300 \"Link to comment\") [^](#comment-1389297 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +3 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I wouldnt call that a \"0.01% bacteria that survives\" solution, the approach is also good even if it can exceed 3n bound in principle, also I admire anyone solving problems in Rust, I wouldnt be able to lol [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n      * | » » » » » [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [show](#) # |\n| ---------------------------------------------------------------------------- | ------------------------ |\n\n| » » » » » [ ![](https://codeforces.com/userpic.codeforces.org/3726532/avatar/64912144a3b5065f.jpg) ](https://codeforces.com/profile/allcaps) [allcaps](https://codeforces.com/profile/allcaps \"Expert allcaps\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389302 \"Link to comment\") [^](#comment-1389300 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Thanks lol. Maybe the type of randomised solution they're talking about cutting is actually something different (tho I can't really think of what it would be). [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n    * | » » » » [Hyder1102](https://codeforces.com/profile/Hyder1102 \"Expert Hyder1102\") | 3 weeks ago, [show](#) # |\n| -------------------------------------------------------------------------------- | ------------------------ |\n\n| » » » » [ ![](https://codeforces.com/userpic.codeforces.org/4156580/avatar/3c7005d8f3eefc09.jpg) ](https://codeforces.com/profile/Hyder1102) [Hyder1102](https://codeforces.com/profile/Hyder1102 \"Expert Hyder1102\") | 3 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389426 \"Link to comment\") [^](#comment-1389297 \"Parent comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +3 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) cool man, apart from randomized u had some nice cost cutting alsothe check +1 possible, if not skip and the upper limit for d [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | -------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n    * | » » » » [abcd1145](https://codeforces.com/profile/abcd1145 \"Expert abcd1145\") | 2 weeks ago, [show](#) # |\n| ----------------------------------------------------------------------------- | ------------------------ |\n\n| » » » » [ ![](https://codeforces.com/userpic.codeforces.org/3337986/avatar/44bb8552ff61c77d.jpg) ](https://codeforces.com/profile/abcd1145) [abcd1145](https://codeforces.com/profile/abcd1145 \"Expert abcd1145\") | 2 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/156316?#comment-1389460 \"Link to comment\") [^](#comment-1389297 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/85295/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) An amazing solution [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [Protorianes](https://codeforces.com/profile/Protorianes \"Specialist Protorianes\") | 3 weeks ago, [show](#) # |\n| ------------------------------------------------------------------------------------ | ------------------------ |\n\n\n...127483 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0adfa9c99001TriRhFYsi5yxYT\nUse the Task tool to have explore agent process this file with Grep and Read (with offset/limit). Do NOT read the full file yourself - delegate to save context.",
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          "callID": "call_cA27bikPviP9BmWjPYEZODAi",
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            "input": {
              "patchText": "*** Begin Patch\n*** Add File: main.cpp\n+#include <bits/stdc++.h>\n+using namespace std;\n+using ll = long long;\n+void setIO() { ios::sync_with_stdio(false); cin.tie(nullptr); }\n+struct Bag {\n+    multiset<ll> l, r;\n+    ll dl = 0, dr = 0;\n+    void insert(ll x) {\n+        if (!l.empty() && x <= *l.rbegin() + dl) l.insert(x - dl);\n+        else r.insert(x - dr);\n+    }\n+    void balance(int k) {\n+        while ((int)l.size() > k) {\n+            auto it = prev(l.end());\n+            r.insert(*it + dl - dr);\n+            l.erase(it);\n+        }\n+        while ((int)l.size() < k) {\n+            auto it = r.begin();\n+            l.insert(*it + dr - dl);\n+            r.erase(it);\n+        }\n+    }\n+    void addAbs(int b) {\n+        int m = l.size() + r.size();\n+        int k = min(m, max(0, -b / 2));\n+        balance(k);\n+        bool middle = b + 2 * k == -1 && !r.empty();\n+        ll x = 0;\n+        if (middle) {\n+            x = *r.begin() + dr;\n+            r.erase(r.begin());\n+        }\n+        dl -= 2;\n+        dr += 2;\n+        if (middle) r.insert(x - dr);\n+    }\n+};\n+int main() {\n+    setIO();\n+    int t; cin >> t;\n+    while (t--) {\n+        int n; cin >> n;\n+        vector<int> a(n), par(n, -1), order{0}, sz(n, 1), id(n), b(n);\n+        vector<ll> cost(n);\n+        vector<vector<int>> g(n);\n+        vector<Bag> bags(n);\n+        for (int &x : a) cin >> x;\n+        for (int i = 1, u, v; i < n; ++i) {\n+            cin >> u >> v; --u; --v;\n+            g[u].push_back(v); g[v].push_back(u);\n+        }\n+        for (int i = 0; i < n; ++i) {\n+            int v = order[i];\n+            for (int u : g[v]) if (u != par[v]) {\n+                par[u] = v; order.push_back(u);\n+            }\n+        }\n+        for (int i = n - 1; i >= 0; --i) {\n+            int v = order[i], heavy = -1;\n+            b[v] = a[v] == 0 ? -1 : a[v];\n+            for (int u : g[v]) if (par[u] == v) {\n+                sz[v] += sz[u]; b[v] += b[u]; cost[v] += cost[u];\n+                if (heavy == -1 || sz[u] > sz[heavy]) heavy = u;\n+            }\n+            id[v] = heavy == -1 ? v : id[heavy];\n+            Bag &cur = bags[id[v]];\n+            for (int u : g[v]) if (par[u] == v && u != heavy) {\n+                Bag &other = bags[id[u]];\n+                for (ll x : other.l) cur.insert(x + other.dl);\n+                for (ll x : other.r) cur.insert(x + other.dr);\n+                other.l.clear(); other.r.clear();\n+            }\n+            if (a[v] == 0) cur.insert(0);\n+            cost[v] += abs(b[v]);\n+            cur.addAbs(b[v]);\n+        }\n+        ll ans = cost[0];\n+        Bag &root = bags[id[0]];\n+        for (ll x : root.l) ans += min(0LL, x + root.dl);\n+        for (ll x : root.r) ans += min(0LL, x + root.dr);\n+        cout << ans << '\\n';\n+    }\n+}\n*** End Patch"
            },
            "output": "Success. Updated the following files:\nA private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1/main.cpp",
            "metadata": {
              "diff": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1/main.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1/main.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1/main.cpp\n@@ -0,0 +1,84 @@\n+#include <bits/stdc++.h>\n+using namespace std;\n+using ll = long long;\n+void setIO() { ios::sync_with_stdio(false); cin.tie(nullptr); }\n+struct Bag {\n+    multiset<ll> l, r;\n+    ll dl = 0, dr = 0;\n+    void insert(ll x) {\n+        if (!l.empty() && x <= *l.rbegin() + dl) l.insert(x - dl);\n+        else r.insert(x - dr);\n+    }\n+    void balance(int k) {\n+        while ((int)l.size() > k) {\n+            auto it = prev(l.end());\n+            r.insert(*it + dl - dr);\n+            l.erase(it);\n+        }\n+        while ((int)l.size() < k) {\n+            auto it = r.begin();\n+            l.insert(*it + dr - dl);\n+            r.erase(it);\n+        }\n+    }\n+    void addAbs(int b) {\n+        int m = l.size() + r.size();\n+        int k = min(m, max(0, -b / 2));\n+        balance(k);\n+        bool middle = b + 2 * k == -1 && !r.empty();\n+        ll x = 0;\n+        if (middle) {\n+            x = *r.begin() + dr;\n+            r.erase(r.begin());\n+        }\n+        dl -= 2;\n+        dr += 2;\n+        if (middle) r.insert(x - dr);\n+    }\n+};\n+int main() {\n+    setIO();\n+    int t; cin >> t;\n+    while (t--) {\n+        int n; cin >> n;\n+        vector<int> a(n), par(n, -1), order{0}, sz(n, 1), id(n), b(n);\n+        vector<ll> cost(n);\n+        vector<vector<int>> g(n);\n+        vector<Bag> bags(n);\n+        for (int &x : a) cin >> x;\n+        for (int i = 1, u, v; i < n; ++i) {\n+            cin >> u >> v; --u; --v;\n+            g[u].push_back(v); g[v].push_back(u);\n+        }\n+        for (int i = 0; i < n; ++i) {\n+            int v = order[i];\n+            for (int u : g[v]) if (u != par[v]) {\n+                par[u] = v; order.push_back(u);\n+            }\n+        }\n+        for (int i = n - 1; i >= 0; --i) {\n+            int v = order[i], heavy = -1;\n+            b[v] = a[v] == 0 ? -1 : a[v];\n+            for (int u : g[v]) if (par[u] == v) {\n+                sz[v] += sz[u]; b[v] += b[u]; cost[v] += cost[u];\n+                if (heavy == -1 || sz[u] > sz[heavy]) heavy = u;\n+            }\n+            id[v] = heavy == -1 ? v : id[heavy];\n+            Bag &cur = bags[id[v]];\n+            for (int u : g[v]) if (par[u] == v && u != heavy) {\n+                Bag &other = bags[id[u]];\n+                for (ll x : other.l) cur.insert(x + other.dl);\n+                for (ll x : other.r) cur.insert(x + other.dr);\n+                other.l.clear(); other.r.clear();\n+            }\n+            if (a[v] == 0) cur.insert(0);\n+            cost[v] += abs(b[v]);\n+            cur.addAbs(b[v]);\n+        }\n+        ll ans = cost[0];\n+        Bag &root = bags[id[0]];\n+        for (ll x : root.l) ans += min(0LL, x + root.dl);\n+        for (ll x : root.r) ans += min(0LL, x + root.dr);\n+        cout << ans << '\\n';\n+    }\n+}\n\n",
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                  "patch": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1/main.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1/main.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IOL4U1/main.cpp\n@@ -0,0 +1,84 @@\n+#include <bits/stdc++.h>\n+using namespace std;\n+using ll = long long;\n+void setIO() { ios::sync_with_stdio(false); cin.tie(nullptr); }\n+struct Bag {\n+    multiset<ll> l, r;\n+    ll dl = 0, dr = 0;\n+    void insert(ll x) {\n+        if (!l.empty() && x <= *l.rbegin() + dl) l.insert(x - dl);\n+        else r.insert(x - dr);\n+    }\n+    void balance(int k) {\n+        while ((int)l.size() > k) {\n+            auto it = prev(l.end());\n+            r.insert(*it + dl - dr);\n+            l.erase(it);\n+        }\n+        while ((int)l.size() < k) {\n+            auto it = r.begin();\n+            l.insert(*it + dr - dl);\n+            r.erase(it);\n+        }\n+    }\n+    void addAbs(int b) {\n+        int m = l.size() + r.size();\n+        int k = min(m, max(0, -b / 2));\n+        balance(k);\n+        bool middle = b + 2 * k == -1 && !r.empty();\n+        ll x = 0;\n+        if (middle) {\n+            x = *r.begin() + dr;\n+            r.erase(r.begin());\n+        }\n+        dl -= 2;\n+        dr += 2;\n+        if (middle) r.insert(x - dr);\n+    }\n+};\n+int main() {\n+    setIO();\n+    int t; cin >> t;\n+    while (t--) {\n+        int n; cin >> n;\n+        vector<int> a(n), par(n, -1), order{0}, sz(n, 1), id(n), b(n);\n+        vector<ll> cost(n);\n+        vector<vector<int>> g(n);\n+        vector<Bag> bags(n);\n+        for (int &x : a) cin >> x;\n+        for (int i = 1, u, v; i < n; ++i) {\n+            cin >> u >> v; --u; --v;\n+            g[u].push_back(v); g[v].push_back(u);\n+        }\n+        for (int i = 0; i < n; ++i) {\n+            int v = order[i];\n+            for (int u : g[v]) if (u != par[v]) {\n+                par[u] = v; order.push_back(u);\n+            }\n+        }\n+        for (int i = n - 1; i >= 0; --i) {\n+            int v = order[i], heavy = -1;\n+            b[v] = a[v] == 0 ? -1 : a[v];\n+            for (int u : g[v]) if (par[u] == v) {\n+                sz[v] += sz[u]; b[v] += b[u]; cost[v] += cost[u];\n+                if (heavy == -1 || sz[u] > sz[heavy]) heavy = u;\n+            }\n+            id[v] = heavy == -1 ? v : id[heavy];\n+            Bag &cur = bags[id[v]];\n+            for (int u : g[v]) if (par[u] == v && u != heavy) {\n+                Bag &other = bags[id[u]];\n+                for (ll x : other.l) cur.insert(x + other.dl);\n+                for (ll x : other.r) cur.insert(x + other.dr);\n+                other.l.clear(); other.r.clear();\n+            }\n+            if (a[v] == 0) cur.insert(0);\n+            cost[v] += abs(b[v]);\n+            cur.addAbs(b[v]);\n+        }\n+        ll ans = cost[0];\n+        Bag &root = bags[id[0]];\n+        for (ll x : root.l) ans += min(0LL, x + root.dl);\n+        for (ll x : root.r) ans += min(0LL, x + root.dr);\n+        cout << ans << '\\n';\n+    }\n+}\n",
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            },
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            },
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            }
          ],
          "editorial": "A subtree's best assignment depends on what its ancestors need. So committing to one locally optimal assignment is too greedy: we need to remember the tradeoff between its sum and its cost.\n\nThe useful trick is that this tradeoff is **convex**. We can store its marginal costs instead of a giant DP table.\n\n## 1. A DP with the right parameter\n\nInitially, replace every zero by $-1$. Let:\n\n- $z_v$ be the number of original zeros in $T_v$;\n- $b_v$ be the subtree sum under this initial assignment;\n- $F_v(k)$ be the minimum total cost of vertices in $T_v$ when exactly $k$ of those zeros are changed to $+1$.\n\nEvery change increases the subtree sum by $2$, so its final sum is $b_v+2k$.\n\nBefore charging vertex $v$ itself, combine its children's DPs and the choice at $v$. A fixed vertex contributes only the option of using zero changes, at cost zero. An originally zero vertex contributes the two options $[0,0]$: leave it at $-1$ or change it to $+1$.\n\nCombining two components means\n\n$$H(k)=\\min_{i+j=k}\\bigl(A(i)+B(j)\\bigr).$$\n\nAfter all components are combined,\n\n$$F_v(k)=H(k)+|b_v+2k|.$$\n\nThis is correct, but storing and merging all these tables directly takes quadratic time.\n\n## 2. Store differences, not tables\n\nRepresent a DP $F$ using its first value $C=F(0)$ and its differences\n\n$$d_i=F(i)-F(i-1),\\qquad 1\\le i\\le m.$$\n\nThen\n\n$$F(k)=C+\\sum_{i=1}^{k}d_i.$$\n\nWe will maintain that the differences are nondecreasing. This is discrete convexity: each additional change becomes no cheaper than the previous one.\n\n### Merging convex DPs\n\nSuppose $A$ and $B$ have sorted difference lists. Choosing index $i$ in $A$ means taking the first $i$ differences from its list; similarly for $B$.\n\nTo use $k$ changes in total, take the $k$ smallest differences from the union of both lists. Because each individual list is sorted, these choices can be made as prefixes. For equal values, break ties by keeping each list's original order.\n\nTherefore, the min-plus convolution has:\n\n- first value $A(0)+B(0)$;\n- difference list equal to the sorted multiset union of both lists.\n\nThe expensive DP merge just became a multiset merge. Much better.\n\nThe vertex's own choice is equally simple: if $a_v=0$, insert one difference equal to zero. Otherwise, insert nothing.\n\n## 3. Adding the absolute-value cost\n\nFor the $i$-th difference, adding $|b+2k|$ contributes\n\n$$\\delta_i=|b+2i|-|b+2(i-1)|.$$\n\nThere are only three possibilities:\n\n- $-2$ if the step ends at a nonpositive sum;\n- $0$ if the step goes from $-1$ to $+1$;\n- $+2$ if the step starts at a nonnegative sum.\n\nThese contributions occur in that order. Since both the old differences and the contributions are nondecreasing, the new differences remain nondecreasing. This proves the convexity invariant, starting from the empty list or the single difference $0$.\n\nIf there are $m$ differences, exactly\n\n$$q=\\min\\left(m,\\max\\left(0,\\left\\lfloor\\frac{-b}{2}\\right\\rfloor\\right)\\right)$$\n\nof them receive $-2$. If $b+2q=-1$ and another difference exists, the next difference receives $0$. All remaining differences receive $+2$.\n\nAlso increase the first DP value by $|b|$.\n\n**Watch the parity:** the step $-1\\to+1$ costs zero. Treating it as $+2$ is a very efficient way to get a wrong answer.\n\n## 4. Two multisets with lazy offsets\n\nMaintain each difference list in a bag containing:\n\n- `l`: a prefix of the sorted list;\n- `r`: the remaining suffix;\n- offsets `dl` and `dr`.\n\nAn element stored as $x$ in `l` has actual value $x+dl$, and similarly for `r`. Every actual value in `l` must be at most every actual value in `r`.\n\n### Insertion and merging\n\nTo insert an actual value $x$, compare it with the largest actual value in `l`. Insert into the appropriate multiset, subtracting that multiset's offset from $x$ first. If `l` is empty, insert into `r`.\n\nTo merge bags, enumerate the source bag's actual values and insert them into the destination. Then clear the source.\n\n### Absolute-value update\n\n1. Move boundary elements between the multisets until `l` contains exactly $q$ elements. Convert between offsets when moving an element.\n2. If the next difference corresponds to $-1\\to+1$, temporarily remove the minimum element of `r`, remembering its actual value.\n3. Decrease `dl` by $2$ and increase `dr` by $2$.\n4. Reinsert the exempted element into `r` with its unchanged actual value.\n\nThe ordering remains valid: a prefix decreases, an optional middle element stays unchanged, and the suffix increases.\n\n## 5. Reuse the largest child's bag\n\nProcess vertices bottom-up. At vertex $v$:\n\n1. Compute its subtree size and baseline sum $b_v$.\n2. Add the children's first DP values.\n3. Reuse the bag of the child with the **largest subtree size**.\n4. Insert all differences from the other children's bags.\n5. Insert $0$ if $a_v=0$.\n6. Add $|b_v|$ to the first DP value and perform the absolute-value update.\n\nUse subtree size to choose the retained child, not just the number of differences. Fixed vertices can shift the required multiset boundary even though they introduce no differences, so they matter for the complexity argument.\n\nAt the root, the DP decreases while its differences are negative and never decreases after that. Hence\n\n$$\\boxed{\\text{answer}=F_1(0)+\\sum_{d<0}d}.$$\n\nZero differences may be taken or skipped; either way, the minimum cost is the same.\n\n## Correctness proof\n\nWe prove by induction over subtrees that each computed bag represents the exact DP $F_v$, with sorted differences.\n\nBefore processing children, a fixed vertex has only the zero-change option, while an originally zero vertex has two zero-cost options. Their difference lists are respectively empty and $[0]$, both sorted.\n\nAssume all children's representations are correct. Merging their sorted differences computes their exact min-plus convolution: the cheapest $k$ increments can be selected as prefixes of the children's sorted lists. Including the vertex's own option therefore gives the minimum descendant cost for every possible number of changes.\n\nThe absolute-value update adds exactly $|b_v+2k|$ to every state: it adds $|b_v|$ to the first value and the corresponding consecutive differences to the list. These added differences are nondecreasing, so the resulting list is sorted and represents exactly $F_v$.\n\nThus the invariant holds at every vertex. At the root, taking all negative differences minimizes the represented DP, so the algorithm outputs the minimum possible total tree cost.\n\n## Complexity and implementation details\n\nCall every child other than the retained largest child a light child. Across a light edge, subtree size at least doubles when moving upward. Thus\n\n$$\\sum_{\\text{light child }u}|T_u|=O(n\\log n).$$\n\nEvery explicitly merged difference can be charged to a vertex in a light subtree, giving $O(n\\log n)$ insertions.\n\nBoundary movements have the same bound. Relative to the retained child's bag, let $W$ be one plus the total size of the light subtrees. The baseline sum changes by at most $W$, at most $W$ new differences are inserted, and the required prefix length changes by $O(W)$. Therefore only $O(W)$ boundary transfers are needed at this vertex. Summing gives $O(n\\log n)$ transfers overall.\n\nEach multiset operation costs $O(\\log n)$, so total time is **$O(n\\log^2 n)$**, with **$O(n)$ memory**.\n\nUse `long long` for costs: a chain with all weights $+1$ has cost $n(n+1)/2$. The implementation builds a parent-before-child order iteratively and processes it backward, avoiding recursion depth issues on a chain. Its integer division in the formula for $q$ is safe: when $b>0$, the outer maximum clamps the result to zero.",
          "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nstruct Bag {\n    multiset<ll> l, r;\n    ll dl = 0, dr = 0;\n\n    void insert(ll x) {\n        if (!l.empty() && x <= *l.rbegin() + dl)\n            l.insert(x - dl);\n        else\n            r.insert(x - dr);\n    }\n\n    void balance(int k) {\n        while ((int)l.size() > k) {\n            auto it = prev(l.end());\n            r.insert(*it + dl - dr);\n            l.erase(it);\n        }\n        while ((int)l.size() < k) {\n            auto it = r.begin();\n            l.insert(*it + dr - dl);\n            r.erase(it);\n        }\n    }\n\n    void addAbs(int b) {\n        int m = l.size() + r.size();\n        int k = min(m, max(0, -b / 2));\n        balance(k);\n\n        // The difference for -1 -> +1 must stay unchanged.\n        bool middle = b + 2 * k == -1 && !r.empty();\n        ll x = 0;\n        if (middle) {\n            x = *r.begin() + dr;\n            r.erase(r.begin());\n        }\n        dl -= 2;\n        dr += 2;\n        if (middle) r.insert(x - dr);\n    }\n};\n\nint main() {\n    setIO();\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<int> a(n), par(n, -1), order{0};\n        vector<int> sz(n, 1), id(n), b(n);\n        vector<ll> cost(n);\n        vector<vector<int>> g(n);\n        vector<Bag> bags(n);\n\n        for (int &x : a) cin >> x;\n        for (int i = 1, u, v; i < n; ++i) {\n            cin >> u >> v;\n            --u;\n            --v;\n            g[u].push_back(v);\n            g[v].push_back(u);\n        }\n\n        for (int i = 0; i < n; ++i) {\n            int v = order[i];\n            for (int u : g[v]) if (u != par[v]) {\n                par[u] = v;\n                order.push_back(u);\n            }\n        }\n\n        for (int i = n - 1; i >= 0; --i) {\n            int v = order[i], heavy = -1;\n            b[v] = a[v] == 0 ? -1 : a[v];\n            for (int u : g[v]) if (par[u] == v) {\n                sz[v] += sz[u];\n                b[v] += b[u];\n                cost[v] += cost[u];\n                if (heavy == -1 || sz[u] > sz[heavy]) heavy = u;\n            }\n\n            id[v] = heavy == -1 ? v : id[heavy];\n            Bag &cur = bags[id[v]];\n            for (int u : g[v]) if (par[u] == v && u != heavy) {\n                Bag &other = bags[id[u]];\n                for (ll x : other.l) cur.insert(x + other.dl);\n                for (ll x : other.r) cur.insert(x + other.dr);\n                other.l.clear();\n                other.r.clear();\n            }\n\n            if (a[v] == 0) cur.insert(0);\n            cost[v] += abs(b[v]);\n            cur.addAbs(b[v]);\n        }\n\n        ll ans = cost[0];\n        Bag &root = bags[id[0]];\n        for (ll x : root.l) ans += min(0LL, x + root.dl);\n        for (ll x : root.r) ans += min(0LL, x + root.dr);\n        cout << ans << '\\n';\n    }\n}"
        },
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          "id": "prt_0adfbc08b001Y1O0Z7yHlRu8xw",
          "sessionID": "ses_f520583c9ffeiTq6VbX8TnWO8h",
          "messageID": "msg_0adfbb5ed0015Vod6jE78V8hLx"
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        {
          "type": "reasoning",
          "text": "",
          "time": {
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            "end": 1789625356136
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          "tool": "StructuredOutput",
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            "input": {
              "status": "success",
              "reason": null,
              "hints": [
                {
                  "order": 1,
                  "content": "Start by assigning every zero $-1$. If $b_v$ is the resulting sum in $T_v$, changing exactly $k$ of its zeros to $+1$ makes that sum $b_v+2k$. Try a subtree DP indexed by $k$."
                },
                {
                  "order": 2,
                  "content": "Let $F_v(k)$ be the minimum cost inside $T_v$ with exactly $k$ zeros changed to $+1$. Combining children is a min-plus convolution, followed by adding $|b_v+2k|$. A plain DP is too slow. What structure do the consecutive differences $F_v(k)-F_v(k-1)$ have?"
                },
                {
                  "order": 3,
                  "content": "Those differences are nondecreasing: the DP is discretely convex. To merge two convex DPs, add their values at index zero and merge their sorted lists of differences. Choosing $k$ increments means taking the $k$ cheapest available differences; their sorted order respects each child's prefix requirement."
                },
                {
                  "order": 4,
                  "content": "Adding $|b+2k|$ changes each difference by $-2$, $0$, or $+2$, in that order. The $0$ occurs only for the step from subtree sum $-1$ to $+1$. Keep the differences in two ordered multisets with lazy offsets, split just before the nonnegative part of this absolute-value contribution."
                },
                {
                  "order": 5,
                  "content": "Reuse the bag of the child with the largest subtree and insert the other children's differences into it. Rebalance the multisets so the left one contains $\\min(m,\\max(0,\\lfloor-b/2\\rfloor))$ differences, where $m$ is their total count. Subtract $2$ lazily on the left and add $2$ on the right, exempting the possible $-1\\to+1$ difference. At the root, add all negative differences to $F_1(0)$. This gives $O(n\\log^2 n)$ time and $O(n)$ space."
                }
              ],
              "editorial": "A subtree's best assignment depends on what its ancestors need. So committing to one locally optimal assignment is too greedy: we need to remember the tradeoff between its sum and its cost.\n\nThe useful trick is that this tradeoff is **convex**. We can store its marginal costs instead of a giant DP table.\n\n## 1. A DP with the right parameter\n\nInitially, replace every zero by $-1$. Let:\n\n- $z_v$ be the number of original zeros in $T_v$;\n- $b_v$ be the subtree sum under this initial assignment;\n- $F_v(k)$ be the minimum total cost of vertices in $T_v$ when exactly $k$ of those zeros are changed to $+1$.\n\nEvery change increases the subtree sum by $2$, so its final sum is $b_v+2k$.\n\nBefore charging vertex $v$ itself, combine its children's DPs and the choice at $v$. A fixed vertex contributes only the option of using zero changes, at cost zero. An originally zero vertex contributes the two options $[0,0]$: leave it at $-1$ or change it to $+1$.\n\nCombining two components means\n\n$$H(k)=\\min_{i+j=k}\\bigl(A(i)+B(j)\\bigr).$$\n\nAfter all components are combined,\n\n$$F_v(k)=H(k)+|b_v+2k|.$$\n\nThis is correct, but storing and merging all these tables directly takes quadratic time.\n\n## 2. Store differences, not tables\n\nRepresent a DP $F$ using its first value $C=F(0)$ and its differences\n\n$$d_i=F(i)-F(i-1),\\qquad 1\\le i\\le m.$$\n\nThen\n\n$$F(k)=C+\\sum_{i=1}^{k}d_i.$$\n\nWe will maintain that the differences are nondecreasing. This is discrete convexity: each additional change becomes no cheaper than the previous one.\n\n### Merging convex DPs\n\nSuppose $A$ and $B$ have sorted difference lists. Choosing index $i$ in $A$ means taking the first $i$ differences from its list; similarly for $B$.\n\nTo use $k$ changes in total, take the $k$ smallest differences from the union of both lists. Because each individual list is sorted, these choices can be made as prefixes. For equal values, break ties by keeping each list's original order.\n\nTherefore, the min-plus convolution has:\n\n- first value $A(0)+B(0)$;\n- difference list equal to the sorted multiset union of both lists.\n\nThe expensive DP merge just became a multiset merge. Much better.\n\nThe vertex's own choice is equally simple: if $a_v=0$, insert one difference equal to zero. Otherwise, insert nothing.\n\n## 3. Adding the absolute-value cost\n\nFor the $i$-th difference, adding $|b+2k|$ contributes\n\n$$\\delta_i=|b+2i|-|b+2(i-1)|.$$\n\nThere are only three possibilities:\n\n- $-2$ if the step ends at a nonpositive sum;\n- $0$ if the step goes from $-1$ to $+1$;\n- $+2$ if the step starts at a nonnegative sum.\n\nThese contributions occur in that order. Since both the old differences and the contributions are nondecreasing, the new differences remain nondecreasing. This proves the convexity invariant, starting from the empty list or the single difference $0$.\n\nIf there are $m$ differences, exactly\n\n$$q=\\min\\left(m,\\max\\left(0,\\left\\lfloor\\frac{-b}{2}\\right\\rfloor\\right)\\right)$$\n\nof them receive $-2$. If $b+2q=-1$ and another difference exists, the next difference receives $0$. All remaining differences receive $+2$.\n\nAlso increase the first DP value by $|b|$.\n\n**Watch the parity:** the step $-1\\to+1$ costs zero. Treating it as $+2$ is a very efficient way to get a wrong answer.\n\n## 4. Two multisets with lazy offsets\n\nMaintain each difference list in a bag containing:\n\n- `l`: a prefix of the sorted list;\n- `r`: the remaining suffix;\n- offsets `dl` and `dr`.\n\nAn element stored as $x$ in `l` has actual value $x+dl$, and similarly for `r`. Every actual value in `l` must be at most every actual value in `r`.\n\n### Insertion and merging\n\nTo insert an actual value $x$, compare it with the largest actual value in `l`. Insert into the appropriate multiset, subtracting that multiset's offset from $x$ first. If `l` is empty, insert into `r`.\n\nTo merge bags, enumerate the source bag's actual values and insert them into the destination. Then clear the source.\n\n### Absolute-value update\n\n1. Move boundary elements between the multisets until `l` contains exactly $q$ elements. Convert between offsets when moving an element.\n2. If the next difference corresponds to $-1\\to+1$, temporarily remove the minimum element of `r`, remembering its actual value.\n3. Decrease `dl` by $2$ and increase `dr` by $2$.\n4. Reinsert the exempted element into `r` with its unchanged actual value.\n\nThe ordering remains valid: a prefix decreases, an optional middle element stays unchanged, and the suffix increases.\n\n## 5. Reuse the largest child's bag\n\nProcess vertices bottom-up. At vertex $v$:\n\n1. Compute its subtree size and baseline sum $b_v$.\n2. Add the children's first DP values.\n3. Reuse the bag of the child with the **largest subtree size**.\n4. Insert all differences from the other children's bags.\n5. Insert $0$ if $a_v=0$.\n6. Add $|b_v|$ to the first DP value and perform the absolute-value update.\n\nUse subtree size to choose the retained child, not just the number of differences. Fixed vertices can shift the required multiset boundary even though they introduce no differences, so they matter for the complexity argument.\n\nAt the root, the DP decreases while its differences are negative and never decreases after that. Hence\n\n$$\\boxed{\\text{answer}=F_1(0)+\\sum_{d<0}d}.$$\n\nZero differences may be taken or skipped; either way, the minimum cost is the same.\n\n## Correctness proof\n\nWe prove by induction over subtrees that each computed bag represents the exact DP $F_v$, with sorted differences.\n\nBefore processing children, a fixed vertex has only the zero-change option, while an originally zero vertex has two zero-cost options. Their difference lists are respectively empty and $[0]$, both sorted.\n\nAssume all children's representations are correct. Merging their sorted differences computes their exact min-plus convolution: the cheapest $k$ increments can be selected as prefixes of the children's sorted lists. Including the vertex's own option therefore gives the minimum descendant cost for every possible number of changes.\n\nThe absolute-value update adds exactly $|b_v+2k|$ to every state: it adds $|b_v|$ to the first value and the corresponding consecutive differences to the list. These added differences are nondecreasing, so the resulting list is sorted and represents exactly $F_v$.\n\nThus the invariant holds at every vertex. At the root, taking all negative differences minimizes the represented DP, so the algorithm outputs the minimum possible total tree cost.\n\n## Complexity and implementation details\n\nCall every child other than the retained largest child a light child. Across a light edge, subtree size at least doubles when moving upward. Thus\n\n$$\\sum_{\\text{light child }u}|T_u|=O(n\\log n).$$\n\nEvery explicitly merged difference can be charged to a vertex in a light subtree, giving $O(n\\log n)$ insertions.\n\nBoundary movements have the same bound. Relative to the retained child's bag, let $W$ be one plus the total size of the light subtrees. The baseline sum changes by at most $W$, at most $W$ new differences are inserted, and the required prefix length changes by $O(W)$. Therefore only $O(W)$ boundary transfers are needed at this vertex. Summing gives $O(n\\log n)$ transfers overall.\n\nEach multiset operation costs $O(\\log n)$, so total time is **$O(n\\log^2 n)$**, with **$O(n)$ memory**.\n\nUse `long long` for costs: a chain with all weights $+1$ has cost $n(n+1)/2$. The implementation builds a parent-before-child order iteratively and processes it backward, avoiding recursion depth issues on a chain. Its integer division in the formula for $q$ is safe: when $b>0$, the outer maximum clamps the result to zero.",
              "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nstruct Bag {\n    multiset<ll> l, r;\n    ll dl = 0, dr = 0;\n\n    void insert(ll x) {\n        if (!l.empty() && x <= *l.rbegin() + dl)\n            l.insert(x - dl);\n        else\n            r.insert(x - dr);\n    }\n\n    void balance(int k) {\n        while ((int)l.size() > k) {\n            auto it = prev(l.end());\n            r.insert(*it + dl - dr);\n            l.erase(it);\n        }\n        while ((int)l.size() < k) {\n            auto it = r.begin();\n            l.insert(*it + dr - dl);\n            r.erase(it);\n        }\n    }\n\n    void addAbs(int b) {\n        int m = l.size() + r.size();\n        int k = min(m, max(0, -b / 2));\n        balance(k);\n\n        // The difference for -1 -> +1 must stay unchanged.\n        bool middle = b + 2 * k == -1 && !r.empty();\n        ll x = 0;\n        if (middle) {\n            x = *r.begin() + dr;\n            r.erase(r.begin());\n        }\n        dl -= 2;\n        dr += 2;\n        if (middle) r.insert(x - dr);\n    }\n};\n\nint main() {\n    setIO();\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<int> a(n), par(n, -1), order{0};\n        vector<int> sz(n, 1), id(n), b(n);\n        vector<ll> cost(n);\n        vector<vector<int>> g(n);\n        vector<Bag> bags(n);\n\n        for (int &x : a) cin >> x;\n        for (int i = 1, u, v; i < n; ++i) {\n            cin >> u >> v;\n            --u;\n            --v;\n            g[u].push_back(v);\n            g[v].push_back(u);\n        }\n\n        for (int i = 0; i < n; ++i) {\n            int v = order[i];\n            for (int u : g[v]) if (u != par[v]) {\n                par[u] = v;\n                order.push_back(u);\n            }\n        }\n\n        for (int i = n - 1; i >= 0; --i) {\n            int v = order[i], heavy = -1;\n            b[v] = a[v] == 0 ? -1 : a[v];\n            for (int u : g[v]) if (par[u] == v) {\n                sz[v] += sz[u];\n                b[v] += b[u];\n                cost[v] += cost[u];\n                if (heavy == -1 || sz[u] > sz[heavy]) heavy = u;\n            }\n\n            id[v] = heavy == -1 ? v : id[heavy];\n            Bag &cur = bags[id[v]];\n            for (int u : g[v]) if (par[u] == v && u != heavy) {\n                Bag &other = bags[id[u]];\n                for (ll x : other.l) cur.insert(x + other.dl);\n                for (ll x : other.r) cur.insert(x + other.dr);\n                other.l.clear();\n                other.r.clear();\n            }\n\n            if (a[v] == 0) cur.insert(0);\n            cost[v] += abs(b[v]);\n            cur.addAbs(b[v]);\n        }\n\n        ll ans = cost[0];\n        Bag &root = bags[id[0]];\n        for (ll x : root.l) ans += min(0LL, x + root.dl);\n        for (ll x : root.r) ans += min(0LL, x + root.dr);\n        cout << ans << '\\n';\n    }\n}"
            },
            "output": "Structured output captured successfully.",
            "metadata": {
              "valid": true
            },
            "title": "Structured Output",
            "time": {
              "start": 1789625457739,
              "end": 1789625457746
            }
          },
          "metadata": {
            "openai": {
              "itemId": "fc_0d837d9ffd5927e7016aab840beb9887d1b26cae5551fdc529"
            }
          },
          "id": "prt_0adfbcf6a001YY360m3WuhihaA",
          "sessionID": "ses_f520583c9ffeiTq6VbX8TnWO8h",
          "messageID": "msg_0adfbb5ed0015Vod6jE78V8hLx"
        },
        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 28904,
            "input": 548,
            "output": 3477,
            "reasoning": 47,
            "cache": {
              "write": 0,
              "read": 24832
            }
          },
          "cost": 0,
          "id": "prt_0adfd5cb8001RXov9zu67AiA9A",
          "sessionID": "ses_f520583c9ffeiTq6VbX8TnWO8h",
          "messageID": "msg_0adfbb5ed0015Vod6jE78V8hLx"
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}
