题目地址：
http://poj.org/problem?id=2524
题目内容：
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 26119 | Accepted: 12859 |
Description
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
Output
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
Hint
Source
#include <stdio.h> int set[];
int n,m; int find_father(int son)
{
if (set[son] == ) {
return son;
}
return set[son] = find_father(set[son]);
} void init()
{
for (int i = ; i < n; i ++) {
set[i] = ;
}
} int main(void)
{
int count = ;
int man1,man2;
while (scanf("%d%d", &n, &m)) {
if (m == && n == )
break;
int res = n;
for (int i = ; i < m; i ++) {
scanf("%d%d", &man1, &man2);
int fat1 = find_father(man1);
int fat2 = find_father(man2);
if (fat1 != fat2) {
set[fat2] = fat1;
res --; // 合并一次，减少一个集合
}
}
printf("Case %d: %d\n", count, res);
init();
count ++;
}
return ;
}