sdut 2428 Ubiquitous Religions（最小生成树）

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本文介绍了一种通过询问学生两两之间的宗教信仰是否相同来间接统计大学校园内不同宗教信仰数量的算法。该算法利用并查集数据结构进行优化，有效地解决了在保护个人隐私的同时获取校园内的宗教多样性的问题。

Ubiquitous Religions

Time Limit: 1000MS Memory limit: 65536K

题目描述

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.

输入

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.

输出

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.

示例输入

示例输出

Case 1: 1
Case 2: 7

#include <stdio.h>
#include <string.h>

int f[50010];

int Find(int x)
{
	return x == f[x] ? x : Find(f[x]);
}

void Link(int x, int y)
{
	int fx = Find(x);
	int fy = Find(y);
	if(fx != fy)
		f[fx] = fy;
}
void Init(int n)
{
	for(int i=1; i<=n; i++)
		f[i] = i;
}

int main()
{
	int n, m;
	int cas = 0;
	
	while(~scanf("%d %d",&n, &m) && (m || n))
	{
		Init(n);
		while(m--)
		{
			int u,v;
			scanf("%d %d",&u, &v);
			Link(u, v);
		}
		int ant = 0;
		for(int i=1; i<=n; i++)
			if(f[i] == i)
				ant++;
		printf("Case %d: %d\n",++cas, ant);
	}
	return 0;
}