A Bug's Life Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 28211   Accepted: 9177 Description Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders…

Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their in…

Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes that they feature two different genders and that they only interact with bugs of the opposite gender. In his experiment, individual bugs and their in…

http://poj.org/problem?id=2492 题意 :就是给你n条虫子,m对关系,每一对关系的双方都是异性的,让你找出有没有是同性恋的. 思路 :这个题跟POJ1703其实差不多,也是需要一个数组来存跟父亲节点的关系,需要两个集合来存是否有关系,在最后稍微变一下形就OK了. #include<iostream> #include<string.h> #include<stdio.h> #include<stdlib.h> using name…

题目链接:http://poj.org/problem?id=2492 A Bug's Life Time Limit: 10000MS   Memory Limit: 65536K Total Submissions: 39415   Accepted: 12835 Description Background Professor Hopper is researching the sexual behavior of a rare species of bugs. He assumes th…

http://blog.csdn.net/freezhanacmore/article/details/8774033?reload  这篇讲解非常好,我也是受这篇文章的启发才做出来的. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 1…

思路:见代码吧. #include <stdio.h> #include <string.h> #include <set> #include <vector> #include <queue> using namespace std; ; int n,m; int father[maxn]; int num[maxn]; //num[i]表示 int rel[maxn]; //rel[i]表示它与根节点的关系,0代表同性,1代表异性 void…

传送门 关键在于到根节点的距离,如果两个点到根节点的距离相等,那么他们性别肯定就一样(因为前面如果没有特殊情况,两个点就是一男一女的).一旦遇到性别一样的,就说明找到了可疑的 #include<bits/stdc++.h> using namespace std; ],n,m,rankk[]; bool flag; inline void init() { flag=false; ; i<=n; ++i) f[i]=i, rankk[i]=; } int getf(int v) { if…

题目链接 这种类型的题目以前见过,今天第一次写,具体过程,还要慢慢理解. #include <cstring> #include <cstdio> #include <string> #include <iostream> #include <algorithm> #include <vector> using namespace std; ]; ]; int find(int x) { if (x == o[x]) return x…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1829 并查集的一个应用,就是检测是否存在矛盾,就是两个不该相交的集合有了交集.本题就是这样,一种虫子有两种性别,每次m次操作,每次给出(a,b),如果a和b是同性别就出现了错误,也就是说出现了判断它有两种性别的错误.我的策略同样是两个集合,用并查集维护两个集合之间的关系.具体证明请看我的上一篇博客,关于这种做法的正确性的证明. 代码如下: #include<bits/stdc++.h> typed…