Description 给定一颗 \(n\) 个顶点的树 \(\text T\),共 \(n-1\) 次断边操作,每次将树分为两部分 \(\text T_1, \text T_2\),求: \[\sum_{(u, v)\in\text E}\left( \sum_{x \text{为} \text T_1 \text{的重心} } x + \sum_{y \text{为} \text T_2 \text{的重心} } y \right) \] 注意,一棵树的重心可能有两个. Hint \(1\l

题目描述 nodgd家里种了一棵树,有一天nodgd比较无聊,就把这棵树画在了一张纸上.另一天nodgd更无聊,就又画了一张.  这时nodgd发现,两次画的顺序是不一样的,这就导致了原本的某一个节点u0在第一幅图中编号为u1,在第二副图中编号为u2.  于是,nodgd决定检查一下他画出的两棵树到底是不是一样的.nodgd已经给每棵树的节点都从1到n进行了编号,即每棵树有n个节点.  如果存在一个1到n的排列p1,p2…pn,对于第一幅图中的任意一条边(i,j),在第二幅图中都能找到一条边(p

Godfather Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6121   Accepted: 2164 Description Last years Chicago was full of gangster fights and strange murders. The chief of the police got really tired of all these crimes, and decided to

http://poj.org/problem? id=1655 Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9072   Accepted: 3765 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a f

树的重心 我们先来认识一下树的重心. 树的重心也叫树的质心.找到一个点,其所有的子树中最大的子树节点数最少,那么这个点就是这棵树的重心,删去重心后,生成的多棵树尽可能平衡. 根据树的重心的定义,我们可以通过树形DP来求解树的重心. 设\(Max_i\)代表删去i节点后树中剩下子树中节点最多的一个子树的节点数.由于删去节点i至少将原树分为两部分,所以满足\(\ \frac{1}{2}n \leq Max_i\),我们要求的就是一个\(i\),使得\(Max_i\)最小. 对于Max数组,我们可以列

Tree Cutting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4834   Accepted: 2958 Description After Farmer John realized that Bessie had installed a "tree-shaped" network among his N (1 <= N <= 10,000) barns at an incredible

Luogu 2680 NOIP 2015 运输计划(树链剖分,LCA,树状数组,树的重心,二分,差分) Description L 国有 n 个星球,还有 n-1 条双向航道,每条航道建立在两个星球之间,这 n-1 条航道连通了 L 国的所有星球. 小 P 掌管一家物流公司,该公司有很多个运输计划,每个运输计划形如:有一艘物 流飞船需要从 ui 号星球沿最快的宇航路径飞行到 vi 号星球去.显然,飞船驶过一条航道 是需要时间的,对于航道 j,任意飞船驶过它所花费的时间为 tj,并且任意两艘飞船之

链接:http://poj.org/problem?id=1655 Time Limit: 1000MS Memory Limit: 65536K Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the ba

Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13178   Accepted: 5565 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or m

Godfather Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7885   Accepted: 2786 Description Last years Chicago was full of gangster fights and strange murders. The chief of the police got really tired of all these crimes, and decided to

题目链接: E. Connecting Universities time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Treeland is a country in which there are n towns connected by n - 1 two-way road such that it's possible to

传送门 Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14070   Accepted: 5939 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one

题目链接: Holiday's Accommodation Time Limit: 8000/4000 MS (Java/Others)     Memory Limit: 200000/200000 K (Java/Others) Problem Description   Nowadays, people have many ways to save money on accommodation when they are on vacation.One of these ways is e

题目链接: Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11845   Accepted: 4993 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of on

Code: #include<bits/stdc++.h> #define maxn 200000 #define inf 1000000000 using namespace std; void setIO(string s) { string in=s+".in",out=s+".out"; freopen(in.c_str(),"r",stdin); } multiset<int>sonmax[maxn]; int

Godfather Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8728   Accepted: 3064 Description Last years Chicago was full of gangster fights and strange murders. The chief of the police got really tired of all these crimes, and decided to

Godfather Time Limit: 2000MS Memory Limit: 65536K Description Last years Chicago was full of gangster fights and strange murders. The chief of the police got really tired of all these crimes, and decided to arrest the mafia leaders. Unfortunately, th

B. Beggin' For A Node time limit per test2.0 s memory limit per test256 MB inputstandard input outputstandard output This is an interactive problem Low_ has a beautiful tree, which he keeps very carefully. A tree is a tree, but mathematically, it cou

description analysis 需要知道一棵树的重心一定在从根出发的重链上,可以考虑先进行树链剖分弄出重儿子和次重儿子,再倍增维护重儿子 由于重链上有一个或两个重心,接下来求的重心都是深度较大的,只需判断其父节点是否也满足重心的性质即可 现在要断掉一条边\((x,y)\),假设\(x\)是\(y\)的父亲,需要分别求出\(y\)的子树的重心.以及除了\(y\)的子树以外的树的重心 倍增数组已经维护好了所以\(y\)的重心很好求,对于视作\(x\)为根的子树则需要重新维护一次倍增数组 若

树的重心定义为:找到一个点,其所有的子树中最大的子树节点数最少,那么这个点就是这棵树的重心,删去重心后,生成的多棵树尽可能平衡. 处理处每个节点的孩子有几个,和树的大小就好了. #include<cstdio> #include<queue> #include<cstring> #include<iostream> #include<algorithm> #define INF 99999999 using namespace std; ; st