POJ-2524Ubiquitous Religions

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该博客介绍了一个竞赛题目，涉及统计大学中不同宗教的数量。由于直接询问每个学生的信仰可能困难，可以通过询问学生对对之间是否信仰相同来推断最多可能存在的宗教数量。博客给出了一个C++实现的解决方案，利用并查集（Union-Find）数据结构来合并属于同一宗教的学生群体，并计算不同的宗教数目。

Ubiquitous Religions

Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 51339 Accepted: 23885

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

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int p[50010]; 
int find(int x)
{
	if(p[x]!=x)
	{
		return p[x]=find(p[x]);
	}
	return p[x];
}
int merge(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
	{
		p[fx]=fy;
		return 1;
	}
	return 0;
}
int main()
{
	int n,m;
	int a,b;
	int ans;
	int count=1;
	
	while(cin>>n>>m,n,m)
	{
		ans=n;
		for(int i=1;i<=n;i++)
		p[i]=i;//初始化 
		for(int i=1;i<=m;i++)
		{
			cin>>a>>b;
			ans-=merge(a,b);
		}
		cout<<"Case "<<count++<<": "<<ans<<endl;
	}
	return 0;
}