POJ 2524 - Ubiquitous Religions （并查集）

最新推荐文章于 2020-01-09 09:19:45 发布

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本文介绍了一个使用并查集算法解决的问题：通过已知某些学生拥有相同宗教信仰的信息，来估算整个大学中不同宗教信仰的最大可能数量。文章提供了一段C语言实现的代码示例，展示了如何通过并查集算法进行分组，并最终确定不同宗教信仰的数量。

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

题意：

给出了m对人说明他们的宗教信仰相同，求n个人有多少不同的宗教信仰。

思路：

并查集模板题。

代码：

#include<stdio.h>
int f[50010]={0},n,m,sum;
void init()
{
	int i;
	for(i=1;i<=n;i++)
		f[i]=i;
	return;
}
int getf(int v)
{
	if(f[v]==v)
		return v;
	else
	{
		f[v]=getf(f[v]);
		return f[v];
	}
}
void merge(int x,int y)
{
	int t1,t2;
	t1=getf(x);
	t2=getf(y);
	if(t1!=t2)
	{
		f[t2]=t1;
	}
	return;
}
int main()
{
	int a,b,i,j,c=1;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		sum=0;
		if(m==0&&n==0)
			break;
		init();
		for(i=1;i<=m;i++)
		{
			scanf("%d%d",&a,&b);
			merge(a,b);
		}
		for(i=1;i<=n;i++)
		{
			if(f[i]==i)
				sum++;
		}
		printf("Case %d: %d\n",c++,sum);
	}
	return 0;
}