poursoul

NetworkTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 749    Accepted Submission(s): 149Problem DescriptionA network administr

题目链接：Magic Number看懂题目，就知道怎么做#include <stdio.h>#include <vector>#include <string.h>#include <algorithm>using namespace std ;typedef long long LL ;typedef pair < int , int > pii ;#define clr( a , x

题目链接：【HDU】5574 Colorful Tree#include <bits/stdc++.h>using namespace std ;typedef long long LL ;typedef pair < int , int > pii ;#define clr( a , x ) memset ( a , x , sizeof a )const int MAXN = 100005

传送门：【HDU】5293 Tree chain problemmy  code:my~~code:/************************************************************************* > File Name: F.cpp > Author: poursoul > Created Time: 2015年07月22

传送门：【HDU】5156 Harry and Christmas tree题目分析：每一种颜色我们分开考虑他们对所有节点的贡献，对于两个颜色同为c的节点u和v（假设u!=v），那么在lca（u，v）的时候我们需要-1，因为在lca（u，v）一直到根的路径上，颜色c只能影响一次。基于此，我们对每种颜色的所有节点按照dfs序排好序，首先每个节点+1，然后对dfs序相邻的两个节点（注意颜色

传送门：【codeforces】494D. Birthday题目分析：首先明确一点，平方和是可以递推的！1.我们第一次dfs，求出从一个点到其子树的所有点的距离的平方和。大体是保存三个变量来递推，设u的子树上点的个数cnt，到子树上点的距离和sum，到子树上点的距离平方和sqr，每次dfs完一个儿子v以后：dp[u].cnt+=dp[v].cntdp[u].sum+=d

传送门：【HDU】3686 Traffic Real Time Query System

传送门：【HDU】4912 Paths on the tree【HDU】4912 Paths on the tree

传送门：【CodeForces】406D Hill Climbing

传送门：【SPOJ】913 Query on a tree II题目分析：

Connections between citiesTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4304    Accepted Submission(s): 1236Problem DescriptionAf

Distance QueriesTime Limit: 2000MS Memory Limit: 30000KTotal Submissions: 8694 Accepted: 3052Case Time Limit: 1000MSDescriptionFarmer John's cows refused to r

How far away ？Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4702    Accepted Submission(s): 1765Problem DescriptionThere are n hou

Nearest Common AncestorsTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 17144 Accepted: 9128DescriptionA rooted tree is a well-known data structure in com

题目链接：Gardener Bo分情况处理add操作即可。#include <stdio.h>#include <string.h>#include <math.h>#include <vector>#include <algorithm>using namespace std ;typedef unsigned int UI ;typedef pair < int , int > pi