poj 2524 Ubiquitous Religions(并查集)

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 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

Sample Output

Case 1: 1
Case 2: 7

描述：当今世界有许多不同的宗教，很难对他们进行追踪。你感兴趣的是，要找出你的大学里有多少个　　　不同的宗教信仰的学生。你知道有n个学生在大学(0 < n <= 50000).你要求每一个学生他们的宗教信仰是不可行的。此外,许多学生不喜欢表达他们的信仰。避免这些问题的一个方法是问m (0 <= m <= n(n-1)/2)对学生,问他们是否相信相同的宗教（例如他们可能知道他们两个参加相同的教堂）。
    从这个数据，你就可以了解到他们的信仰，但你可以得到一个想法的上限，有多少不同的宗教可以在校园里可能代表。你可以假设每个学生最多信仰一种宗教。

输入：输入包含多组数据。每个案例开始一行指定整数n和m。下面m行包含两个整数i和j,表示学生i和j信仰相同的宗教。学生编号1到n。最后一行中输入指定的n = m = 0。

输出：对于每个测试用例,打印一行数字编号(从1开始),其次是学生们在大学里所信仰不同宗教的最大数量。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
#define MAX 50005
int path[MAX];
void f(int n)
{
    for(int i=1; i<=n; i++)
    {
        path[i]=i;　//初始化为自己是自己的父节点
    }
}
int find(int x) //路径压缩
{
    if(x==path[x])
        return path[x];
    return path[x]=find(path[x]);
}
int main()
{
    int n,m,k=0;
    int a,b,x,y;
    int ans;
    while(cin>>n>>m,n||m)
    {
        f(n);
        ans=n; //初始化宗教最多为n个
        for(int i=1; i<=m; i++)
        {
            cin>>a>>b;
            x=find(a);
            y=find(b);
            if(x!=y) // 两者的父节点不同
            {
                path[y]=x; //连接
                ans--;     //同一信仰，就减一
            }
        }
        printf("Case %d: ",++k);
        cout<<ans<<endl;
    }
    return 0;
}