POJ 2524 （Ubiquitous Religions）

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.

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.

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.

10 9
1 2
1 3
1 4
1 5
1 6
1 7
1 8
1 9
1 10
10 4
2 3
4 5
4 8
5 8
0 0

Case 1: 1
Case 2: 7

思路：没什么好说的 就是并查集.
代码如下:

#include <stdio.h>
int s;
int p[50050];
int find(int x)//找老大. 
{
    if(x!=p[x])
    p[x]=find(p[x]);
    return p[x];
}
void Union (int x,int y)//合并成同一派的老大唯一.不是老大就是老二! 
{
    int fx=find(x),fy=find(y);
    if(fx!=fy)
    {
    s--;//合并一次就减一个帮派. 
    p[y]=p[x];//原来的老大变成现在的老二,小弟都变成老二 
    }
    else
    return;
}
int main()
{
    int i,j;
    int s1,s2;
    int x,y,n,m;
    int t = 1;
    while(~scanf("%d %d",&n,&m))
    {
        if(n==0&&m==0) 
        break;
        s=n;//开始我们有n个帮派. 
        for(i = 1; i <= n; i++)
            p[i] = i;
        for(i = 1; i <= m; i++)
        {
            scanf("%d %d",&x,&y);
            s1 = find(x);
            s2 = find(y);
            Union(s1,s2);
        }
        printf("Case %d: ",t);
        printf("%d\n",s);
      t++;
    }
    return 0;
}