并查集——Ubiquitous Religions

Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 22821		Accepted: 11254

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

Hint

Huge input, scanf is recommended.

题意：

调查学校里的学生信仰情况，给出的是两个人的信仰相同，计算最多有多少种信仰。

并查集的知识已经在上一篇博客中写过了，现在直接贴代码：

#include 
#include 
#include 
#include 

int bj[50009];

int fin(int x)
{
    int r;
    r = x;
    while(bj[r] != r)//查找根节点，只有下表和内容相同的才会是根节点
        r = bj[r];
    return r;
}

void bing(int x,int y)
{
    int fx,fy;
    fx = fin(x);//查找x的根节点
    fy = fin(y);//查找y的根节点
    if(fx != fy)//如果x的根节点和y的根节点不相同，则合并两个集合
    {
        bj[fx] = fy;//将x所在的树合并到y所在的树下，也就是把x的根节点改成y的根节点
    }
}
int main()
{
    int n,m,i,j,k,x,y,t = 0;
    while(~scanf("%d%d",&n,&m))
    {
        if(n == 0 && m == 0)
            break;
        t++;
        for(i = 0; i <= n; i++)
            bj[i] = i;
        for(i = 0; i < m; i++)
        {
            scanf("%d%d",&x,&y);
            bing(x,y);
        }
        int num = 0;
        for(i = 0; i < n; i++)
        {
            if(bj[i] == i)//计算在这个并查集森林中，一共有多少棵树
                num++;
        }
        printf("Case %d: %d\n",t,num);
    }
    return 0;
}

扩展：

如果此题目是询问两个人的信仰是否相同，这就需要对并查集进行根节点的查询，下面展示普通的以及优化的代码：

//这是普通的查询
int find(int x)//两次调用此函数，分别传入x,y，返回他们的根节点，在主函数中进行判断，如果相同，则信仰相同。
{
    int r = x;
    while(bj[r] != r)
        r = bj[r];
    return r;
}

//但是如果查询的次数达到上万次，则就有可能会超时，下面是优化过的代码
int find(int x)
{
    int r = x;
    while(bj[r] != r)
        r = bj[r];
    int k = x,j;
    while(k != r)//按照上次的顺序再次遍历一遍，把每个子节点存储的值都变成根节点存储的值。
    {
        j = bj[k];
        bj[k] = r;
        k = j;
    }
    return r;
}
//这样，只需要判断两个节点下存储的值是不是相同就可以了。个人认为，在建立关系的时候最好就这么办。