poj 2524 Ubiquitous Religions 【并查集】

poj 2524 Ubiquitous Religions 【并查集】

Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 12445		Accepted: 5900

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 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

Sample Output

Case 1: 1
Case 2: 7

第一个并查集程序，最小生成树不算。

 n个点，给你m条边，求最大能有多少个连通分量。

#include < iostream >
using   namespace  std;
const   int  MAX = 50001 ;
int  fa[MAX];

int  find( int  x)
{
     return  fa[x] == x ? x:find(fa[x]);
}

void  Union( int  x,  int  y)
{
     fa[find(x)] = find(y);
}
int  main()
{
     int  n,m;
     for ( int  tt = 1 ; ; tt ++ )
    { 
              cin >> n >> m;
              if ( n == 0 && m == 0 ) break ;
             
              for ( int  i = 1 ; i <= n; i ++ )
                     fa[i] = i; 
                     
              int  max = n;              
              for ( int  i = 1 ,s,t; i <= m; i ++ )
                     {
                         cin >> s >> t;
                          if (find(s) != find(t))max = max - 1 ;
                         Union(s,t);              
                     }
                     
             cout << " Case  " << tt << ' : ' << '   ' << max << endl;
             
    }
    
    system( " pause " );    
     return   0 ;
}