(使用树结构来支持并查集操作8.4.4)POJ 2524 Ubiquitous Religions(并查集的基本操作: 求有多少个相互独立的集合)

/*
 * POJ_2524.cpp
 *
 *  Created on: 2013年11月6日
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>

using namespace std;


const int maxn = 50010;

int father[maxn];

int find(int x){//并查集的查询
	if(father[x] == x){
		return x;
	}

	return (father[x] = find(father[x]));
}


void join(int x,int y){//并查集的合并
	x = find(x);
	y = find(y);

	if(x != y){
		father[x] = y;
	}
}


int main(){
	int n,m;

	int counter = 1;
	while(scanf("%d%d",&n,&m)!=EOF,n||m){
		int x,y;

		int i;
		for(i = 1 ; i <= n  ; ++i){
			father[i] = i;
		}

		for(i = 1 ; i <= m ; ++i){
			scanf("%d%d",&x,&y);

			join(x,y);
		}

		int cnt = 0;
		for(i = 1 ; i <= n ; ++i){//就算有多少个相互独立的集合
			if(father[i] == i){
				cnt++;
			}
		}

		printf("Case %d: %d\n",counter++,cnt);
	}

	return 0;
}