Ubiquitous Religions(简单并查集 poj 2524 )

Ubiquitous Religions

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 15791 Accepted: 7606

Description

There are so many different religions in the world today that it is difficult to keep track of them all. You are interested in finding out how many different religions students in your university believe in. 

You know that there are n students in your university (0 < n <= 50000). It is infeasible for you to ask every student their religious beliefs. Furthermore, many students are not comfortable expressing their beliefs. One way to avoid these problems is to ask m (0 <= m <= n(n-1)/2) pairs of students and ask them whether they believe in the same religion (e.g. they may know if they both attend the same church). From this data, you may not know what each person believes in, but you can get an idea of the upper bound of how many different religions can be possibly represented on campus. You may assume that each student subscribes to at most one religion.

Input

The input consists of a number of cases. Each case starts with a line specifying the integers n and m. The next m lines each consists of two integers i and j, specifying that students i and j believe in the same religion. The students are numbered 1 to n. The end of input is specified by a line in which n = m = 0.

Output

For each test case, print on a single line the case number (starting with 1) followed by the maximum number of different religions that the students in the university believe in.

Sample Input

10 91 21 31 41 51 61 71 81 91 1010 42 34 54 85 80 0

Sample Output

Case 1: 1Case 2: 7

Hint

Huge input, scanf is recommended.

翻译：

当今世界有很多不同的宗教,很难通晓他们。你有兴趣找出在你的大学里有多少种不同的宗教信仰。你知道在你的大学里有n个学生(0 < n <= 50000) 。你无法询问每个学生的宗教信仰。此外,许多学生不想说出他们的信仰。避免这些问题的一个方法是问m(0 <= m <= n(n - 1)/ 2)对学生, 问他们是否信仰相同的宗教( 例如他们可能知道他们两个是否去了相同的教堂) 。在这个数据中,你可能不知道每个人信仰的宗教，但你可以知道校园里最多可能有多少个不同的宗教。假定每个学生最多信仰一个宗教。

简单并查集应用。

#include<cstdio>#include<cstring>const int maxn = 50000 + 5;int root[maxn];int Find(int x)//逐级找老板{int top;if (root[x] == x)//发现自己是自己的老板{return x;}top = Find(root[x]);root[x] = top; //路径压缩return top;}int main(){int a, b, m, n, t = 1,tmp1,tmp2;while (scanf("%d%d",&n,&m)==2&&m){for (int i = 1; i <= n; i++)//把自己设为自己的老板{root[i] = i;}for (int i = 0; i < m; i++){scanf("%d%d", &a, &b);tmp1=Find(a);tmp2 = Find(b);if ( tmp1!=tmp2 )//查找自己的老板是否是同一个{root[tmp1] = tmp2;//公司合并，让其中一个老板做另一个老板的下属n--;}}printf("Case %d: %d\n", t++, n);}return 0;}

转载地址：http://www.cnblogs.com/HDMaxfun/p/5700716.html