Ubiquitous Religions 并查集

最新推荐文章于 2020-01-09 09:19:45 发布

原创最新推荐文章于 2020-01-09 09:19:45 发布 · 138 阅读

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本文介绍了一种通过询问学生是否属于同一宗教来估算大学校园内不同宗教信仰数量的算法。利用并查集数据结构，该算法能够高效地计算出可能存在的最大宗教数量。

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

题意：

知道有n个人，m行的x和y是同一宗教

问学校有几个宗教

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
int a[50005];
int find(int x){
    if(a[x]==x) return x;
    else return a[x]=find(a[x]);
}
void link(int x,int y){
    x=find(x);
    y=find(y);
    if(x!=y){
        a[y]=x;
    }
}


int main()
{
    int n,m;
    int t=1;
    while (scanf("%d%d", &n, &m) && n != 0 && m != 0){
        for(int i=1;i<=n;i++){
            a[i]=i;
        }
        int x,y;
        for(int i=0;i<m;i++){
            scanf("%d %d",&x,&y);
            link(x,y);
        }
        int s=0;
        for(int i=1;i<=n;i++){
            if(a[i]==i) s++;
        }
        printf("Case %d: %d\n",t++,s);
    }

return 0;
}