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Codeforces Round 525 (Div2)

Posted on 2018-12-12 | In ACM , codeforces

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题目链接

A,B,C

签到


D 交互题


E

题意

给一棵点权(可正可负)树,选择任意个不重叠的连通子图,使得

题解

# ACM # codeforces
Codeforces Round 344 (Div2)
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Kzpx

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  1. 1. 题目链接
  • A,B,C
  • D 交互题
  • E
    1. 题意
    2. 题解
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