题目链接:https://cn.vjudge.net/contest/277955#problem/D 题目大意:求树的重心(树的重心指的是树上的某一个点,删掉之后形成的多棵树中节点数最大值最小). 具体思路:对于每一个点,我们求出以当前的点为根的根数的节点个数, 然后在求树的重心的时候,一共有两种情况,一种树去除该点后这个点的子节点中存在所求的最大值,还有一种情况是这个点往上会求出最大值,往上的最大值就是(n-dp[rt][0]). AC代码: #include<iostream> #inc…

Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14550   Accepted: 6173 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…

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…

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…

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…

题意:就是裸的求树的重心. #include<cstring> #include<algorithm> #include<cmath> #include<cstdio> #include<iostream> #define N 20007 #define inf 100000007 using namespace std; int n,id,mnum; int siz[N]; ],rea[N*]; void add(int u,int v) {…

//树形DP+树状数组 HDU 5877 Weak Pair // 思路:用树状数组每次加k/a[i],每个节点ans+=Sum(a[i]) 表示每次加大于等于a[i]的值 // 这道题要离散化 #include <bits/stdc++.h> using namespace std; #define LL long long typedef pair<int,int> pii; const double inf = 123456789012345.0; const LL MOD…

[HDU 5293]Tree chain problem(树形dp+树链剖分) 题面 在一棵树中,给出若干条链和链的权值,求选取不相交的链使得权值和最大. 分析 考虑树形dp,dp[x]表示以x为子树的最大权值和(选的链都在i的子树中) 设sum[x]表示x的儿子的dp值和,即\(\sum _{y \in \mathrm{son}(x)} dp[y]\) 1.不选两端点lca为x的链,dp[x]=sum[x] 2.选两端点lca为x的链,则dp[x]=max{链的权值+链上节点的所有子节点dp的…

Walking Race Time Limit: 10000MS   Memory Limit: 131072K Total Submissions: 4123   Accepted: 1029 Case Time Limit: 3000MS Description flymouse’s sister wc is very capable at sports and her favorite event is walking race. Chasing after the championshi…

Godfather poj-3107 题目大意:求树的重心裸题. 注释:n<=50000. 想法:我们尝试用树形dp求树的重心,关于树的重心的定义在题目中给的很明确.关于这道题,我们邻接矩阵存不下,用链式前向星存边,然后对于任选节点遍历,然后在回溯是进行最大值的最小值更新,之后就是一点显然的结论——树最多只有两个重心,而且这两个加点必须连边. 最后,附上丑陋的代码... ... #include <iostream> #include <cstdio> #include &l…