poj 2524 Ubiquitous Religions_不见不散的结局是曲终人散_新浪博客

Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 27748		Accepted: 13594

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

题意：其中一个学校有N个人，没有人都有宗教信仰，其中同一组的代表宗教信仰相同，求最多有多少不同的宗教信仰；

思路：直接并查集，用一个SUM记录，SUM初始值为N，当遇到相同的宗教信仰SUM--；

代码如下：

#include< stdio.h >

int sum,n;

int f[50005];

int find(int x)

{

    if(f[x]!=x)

        f[x]=find(f[x]);

    return f[x];

}

void panduan(int a,int b)

{

    a=find(a);

    b=find(b);

    if(a==b)//之前此宗教出现过，已经减过；

        return ;

    sum--;

        f[a]=b;

}

int main()

{

    int y,cnt=0;

    while(scanf("%d%d",&n,&y)&&(n||y))

    {

        sum=n;

        while(n--)

            f[n]=n;

        while(y--)

        {

            int c,d;

            scanf("%d%d",&c,&d);

            panduan(c,d);

        }

        printf("Case %d: %d\n",++cnt,sum);

    }

    return 0;

}