Ubiquitous Religions 【poj-2524】【并查集】

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本文介绍了一种使用并查集算法解决宗教信仰调查问题的方法。通过询问学生间的宗教信仰是否相同来估算校园内不同宗教信仰的最大数量。文章提供了一个完整的C++实现案例，并解释了如何初始化、查找和合并节点。

Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 36790		Accepted: 17517

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

Sample Output

Case 1: 1
Case 2: 7

题解：并查集，模板题。

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=50005;
int par[maxn];    //存父亲节点
int rank[maxn];   //存树的高度

void init(int n){    //初始化
	for(int i=1;i<=n;i++){
		par[i]=i;
		rank[i]=0;
	}
}
int find(int x){    //查询树的根
	if(par[x]==x){
		return x;
	}else{
		return par[x]=find(par[x]);
	}
}
void unite(int x,int y){   //合并
	x=find(x);
	y=find(y);
	if(x==y) return ;
	if(rank[x]<rank[y]){
		par[x]=y;
	}else{
		par[y]=x;
		if(rank[x]==rank[y]) rank[x]++;
	}
}

int main()
{
	int n,m;
	int k=1;
	while(scanf("%d%d",&n,&m)){
		if(n==0&&m==0) break;
		init(n);
		int a,b;
		for(int i=0;i<m;i++){
			scanf("%d%d",&a,&b);
			unite(a,b);
		}
		int num=0;
		for(int i=1;i<=n;i++){
			if(par[i]==i) num++;
		}
		printf("Case %d: %d\n",k++,num);
		
	}
	
	
	return 0;
}