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…

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…

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…

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…

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…

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…

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…