#include <stdio.h> #include <string.h> ], g[]; int find(int x) //并查集的查找,找到共同的父亲 { if (f[x] != x) f[x] = find(f[x]); return f[x]; } int main() { ; while (scanf("%d%d", &a, &b) != EOF) { , t = ; && b<) break; ; i <…

Is It A Tree? Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 28838   Accepted: 9843 ->  Link  poj  <- ->  Link  hdu<- ->  Link  NYoj <- 这三个题题目都是一样的,我先做的NYOJ上的,以-1.-1结尾,然后跑到HDu上交一遍跪了,改成同时小于0就A了, 又到POJ上交一遍又跪了,改成-1.-1结尾又A了: 题意…

Is It A Tree? Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14197    Accepted Submission(s): 3170 Problem Description A tree is a well-known data structure that is either empty (null, void, n…

A Bug's Life Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 13709    Accepted Submission(s): 4449 Problem Description Background Professor Hopper is researching the sexual behavior of a rare…

[BZOJ2959]长跑(Link-Cut Tree,并查集) 题面 BZOJ 题解 如果保证不出现环的话 妥妥的\(LCT\)傻逼题 现在可能会出现环 环有什么影响? 那就可以沿着环把所有点全部走一遍吧 所以,相当于把环看成一个点来搞一搞 所以,维护一个并查集 记录一下每个点被缩成了哪个点 然后再用\(LCT\)维护缩点后的树就行啦 #include<iostream> #include<cstdio> #include<cstdlib> #include<cs…

http://poj.org/problem?id=2236 由于发生了地震,有关组织组把一圈电脑一个无线网,但是由于余震的破坏,所有的电脑都被损坏,随着电脑一个个被修好,无线网也逐步恢复工作,但是由于硬件的限制,一台电脑和另一台电脑能够相连当他们之间的距离小于d,或者还有一台电脑当中介,分别与两台电脑相连. 在修复的过程中,工作者会有两种操作,修复电脑和询问电脑a和电脑b是否相连.当询问的时候输出答案. 因为输入数据很大,需要快速判断电脑a和电脑b相连,所以自然想到用并查集. 初始时候 全部电…

Description A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following properties. There is exactly one node, called the root, t…

Is It A Tree? 题目链接:http://poj.org/problem?id=1308 Description: A tree is a well-known data structure that is either empty (null, void, nothing) or is a set of one or more nodes connected by directed edges between nodes satisfying the following proper…

题目链接  Imbalance Value of a Tree 题意  给定一棵树.求树上所有简单路径中的最大权值与最小权值的差值的和. 首先考虑求所有简单路径中的最大权值和. 对所有点按照权值大小升序排序,即若$a[i] < a[j]$,那么$i$排在$j$前面. 接下来开始依次处理.对于每个点$i$,寻找周围跟他连通并且权值比他小的点进行合并,并且累加答案. 整个过程用并查集维护. 那么类似地,求最小权值的和的时候,我们把所有点权变成原来的相反数,再做一遍即可. #include <bit…

Problem Description Luxer is a really bad guy. He destroys everything he met.  One day Luxer went to D-city. D-city has N D-points and M D-lines. Each D-line connects exactly two D-points. Luxer will destroy all the D-lines. The mayor of D-city wants…