poj 2524 Ubiquitous Religions

Ubiquitous Religions Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 29591   Accepted: 14298 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 Hint Huge input, scanf is recommended. Source Alberta Collegiate Programming Contest 2003.10.18

大意：

每一组数据中有m对数据，每一对数据代表两个人信仰一样，求最多有多少种信仰、

           并查集做，合并之后判断父节点的个数

AC代码：

#include<iostream>
using namespace std;
#define size  50010
int father[size];
int find(int x){
	while(father[x]!=x)
	x=father[x];
	return x;
}
void merage(int x,int y){
	x=find(x);
	y=find(y);
	if(x!=y)
	father[x]=y;	
}
int main()
{
	int m,n,i,s=0;
	int x,y;
	while(cin>>n>>m&&(m||n))
	{
		int max=0;
		s++;
		for(i=1;i<=n;i++)
		father[i]=i;
		for(i=0;i<m;i++){
			scanf("%d %d",&x,&y);
	    	merage(x,y);
		}
		for(i=1;i<=n;i++)
		if(father[i]==i)	
		max++;
		printf("Case %d: %d\n",s,max);
	}
	return 0;
} 

也可以进行路径压缩和按秩合并，但是个人觉得不用的话效率会更高点