POJ - 2524：Ubiquitous Religions

Ubiquitous Religions

来源：POJ

标签：并查集

参考资料：

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题目

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.

输入样例

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

输出样例

Case 1: 1
Case 2: 7

题目大意

在学生群体中计算最多存在多少种不同的宗教信仰。给定n代表学生个数，从1开始编号；给定m代表有m种条件，每个条件中给定2个学生编号i和j，代表学生i和学生j的宗教信仰相同。注：每个同学只有1个信仰。

解题思路

利用并查集思想。假定n个学生最多有n种不同的信仰，在每种条件下，判断同学i和j是否已存在于“一个由有着相同信仰的同学组成的集合”，如果不存在，则应加入这个集合，同时不同信仰数减一。

参考代码

#include
#define MAXN 50000
int p[MAXN+5];

int find(int x){
	if(p[x]==x){
		return x;
	}
	return p[x]=find(p[x]);
}

void join(int x,int y){
	int fx=find(x);
	int fy=find(y);
	p[fx]=fy;
}

int has_exist(int x,int y){
	return find(x)==find(y);
}

int main(){
	int t=0;//case计数 
	int n,m;
	while(scanf("%d%d",&n,&m) && n){
		int ans=n;//不同信仰数初始化为n 
		int x,y;//同学x和同学y 
		for(int i=1;i<=n;i++){//假定每个同学都有一种不同的信仰 
			p[i]=i;
		}
		for(int i=1;i<=m;i++){
			scanf("%d%d",&x,&y);
			if(has_exist(x,y)==0){
				ans--;
				join(x,y);
			}
		}
		printf("Case %d: %d\n",++t ,ans);
	}
	return 0;
}