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…

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…

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…

http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=45522 给定一棵树的n个节点,每个节点标号在1到n之间,1是树的根节点,有如下两种操作: M v :把编号为v的节点标记. Q v :查询与v节点距离最近的被标记的点的编号.最初只有根节点被标记. 输入n-1个数表示,每一个数是当前第i个数的父节点,输出对于每个查询得到标号的总和. 并查集与树的结合,在输入的时候就可以根据父节点的关系建立并查集,1的父亲是1. 还有把编号为…

题目链接:http://poj.org/problem?id=2236 题目大意:城市网络由n台电脑组成,因地震全部瘫痪,现在进行修复,规定距离小于等于d的电脑修复之后是可以直接相连 进行若干操作,O a,修复编号为a的电脑,S a,b  询问a,b电脑能否联系 思路分析:并查集,只是和并条件变了,首先要已经被修复(vis数组)其次距离要小于d,并查集搞完之后 询问的时候只需要看他们是不是有着相同的根节点即可. 代码: #include <iostream> #include <cstd…

An earthquake takes place in Southeast Asia. The ACM (Asia Cooperated Medical team) have set up a wireless network with the lap computers, but an unexpected aftershock attacked, all computers in the network were all broken. The computers are repaired…

Ivan had string s consisting of small English letters. However, his friend Julia decided to make fun of him and hid the string s. Ivan preferred making a new string to finding the old one. Ivan knows some information about the string s. Namely, he re…

#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 <…

HDU1213How Many Tables Problem Description Today is Ignatius' birthday. He invites a lot of friends. Now it's dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all…

#include <iostream> #include <algorithm> using namespace std; + ; int fa[maxn]; int Find(int x){ if (x == fa[x]) return x; else return fa[x] = Find(fa[x]); } int main(){ int a, b; while (cin >> a){ //init ; i < maxn; i++) fa[i] = i; ;…