poj 2524 Ubiquitous Religions

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

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.

题意：在学校中有n（从1到n编号）个学生，你问了m对他们是不是信仰同一个宗教。最后输出在这群大学生中所信仰的不同的宗教的数目。n=m=0时结束。

思路：信仰同一个宗教的学生组成一个集合，我们用并查集计算集合数。初始时n个学生各组成一个集合，集合的代表元素就是它本身。以后每输入一对信仰同一宗教的学生（x，y），x集合并入y集合，即y所在集合的代表元素就是x所在集合的代表元素。最后统计有多少个学生其代表元素是其本身就行了

#include <iostream>#include <string.h>using namespace std;int set[50010];int set_find(int x)  //查找元素x所在集合的代表元素{    if(set[x]<0)        return x;    return set[x]=set_find(set[x]);}void set_join(int p,int q)  //合并信仰同一个宗教的学生{    p=set_find(p);    q=set_find(q);    if(p!=q)        set[p]=q;}int main(){    int n,m,loop=0;    while(cin>>n>>m)    {        if(n==0&&m==0)            break;        memset(set,-1,sizeof(set));        for(int t=1;t<=m;t++)        {            int i,j;            cin>>i>>j;            set_join(i,j);        }        int count=0,i;   //统计有多少个宗教        for(i=1;i<=n;i++)            if(set[i]<0)                count++;        cout<<"Case "<<++loop<<": "<<count<<endl;    }    return 0;}