poj2524 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

题目大意：n个人，每个人的宗教信仰可能不同。数据给出一对数x，y ，表示两人第宗教信仰相同。求一共有几种宗教
思路：裸的并查集就好了

/*
 * Author:  Joshua
 * Created Time:  2014年07月11日 星期五 15时16分58秒
 * File Name: poj2524.cpp
 */
#include<cstdio>
#define maxn 50005
typedef long long LL;
int f[maxn];
int n,m,t=0;

int gf(int x)
{
    return f[x]==x ? x : f[x]=gf(f[x]);
}

void solve()
{
    int x,y,temp;
    for (int i=1;i<=n;++i)
        f[i]=i;
    for (int i=1;i<=m;++i)
    {
        scanf("%d%d",&x,&y);
        if (x>y)
        {
            temp=x;x=y;y=temp;
        }
        f[gf(y)]=gf(x);
    }
    int ans=0;
    for (int i=1;i<=n;++i)
       if (f[i]==i) ans++;
    t++;
    printf("Case %d: %d\n",t,ans);
}

int main()
{
    while (scanf("%d%d",&n,&m) && !(n==0 && m==0) )
    {
        solve();
    }
    return 0;
}