Ubiquitous Religions POJ - 2524

 

题目描述：

 

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.

 

 

 

Outout

 

 

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

 

 

#include
#include
using namespace std;
int a[500000];

inline int search_root(int x)
{
    return x==a[x]?x:(search_root(a[x]));//递归查找到根 
}
int main(void)
{
    int m,n;
    int student1,student2;
    int cas=0;
    int religions;
    while(cin>>m>>n)
    {
        if(!m&!n)break;
        religions=m;   //一开始假设所有的人的宗教信仰都不相同 
        for(int i=1;i<=m;i++)  a[i]=i;//初始化,假设每个人宗教编号都不一样 
        for(int i=0;i>student1>>student2;
        
            int root_point1=search_root(a[student1]);
            int root_point2=search_root(a[student2]);

            if(root_point1!=root_point2)
            {
                a[root_point2]=root_point1;
                religions--;//每一次并集都会减少一个宗教数 
            }
        
        }
        cout<<"Case "<<++cas<<": "<