POJ2534 Ubiquitous Religions

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 91 21 31 41 51 61 71 81 91 1010 42 34 54 85 80 0

Sample Output

Case 1: 1Case 2: 7

Hint

Huge input, scanf is recommended.

这道题意思说，每行两个数据，后一个学生信仰的宗教和前一个同学的宗教相同，让你输出总共多少个不同的宗教。

解题思路：首先，将各自的祖先设置为自己本身，并将自己所在家族的成员个数设置为初始值1

大致就是简单的并查集，每输入一组样例，判断他们的祖先是否相同

1.相同，则进入下一组样例的判断。

2.不同，将一个的祖先设置为另外一个庞大家族的祖先。

StatusAcceptedTime297msMemory556kBLength616LangC++Submitted2017-04-18 13:13:24SharedRemoteRunId16895025

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;int a[50007];int s[50007];int num=0;int find(int i){while(i!=a[i])i=a[i];return i;}int main(){int n,m;int stu1,stu2,p,q;while(scanf("%d%d",&n,&m)!=EOF&&(n||m)){for(int i=0;i<n;i++){a[i]=i;s[i]=1;}for(int i=0;i<m;i++){scanf("%d%d",&stu1,&stu2);p=find(stu1);q=find(stu2);if(p!=q){if(s[p]>=s[q]){s[p]+=s[q];a[q]=p;}else{s[q]+=s[p];a[p]=q;}n--;}}printf("Case %d: %d\n",++num,n);}return 0;}