Ubiquitous Religions（并查集）POJ - 2524

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.

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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.

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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.

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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

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Sample Output

Case 1: 1
Case 2: 7

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题意：当今世界有太多不同的宗教，很难把它们都记录下来。你想知道你大学里有多少不同宗教的学生信仰宗教。
你知道你们学校有n个学生（0<n<=50000）。你不可能问每个学生他们的宗教信仰。此外，许多学生不愿意表达他们的信仰。
避免这些问题的一种方法是询问m（0<=m<=n（n-1）/2）对学生，并询问他们是否信仰同一宗教（例如，他们可能知道他们是否都参加了同一个教堂）。
从这些数据中，你可能不知道每个人信仰什么，但你可以了解到有多少不同的宗教可以在校园里代表。你可以假设每个学生最多只信奉一种宗教。

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//看了一位大佬的博客，秒懂并查集，初学可以去看看，写的太好了
//附地址：https://blog.csdn.net/niushuai666/article/details/6662911
#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include <sstream>
#include<vector>
#include<cmath>    
#include<stack>
#include<time.h>
#include<ctime>
using namespace std;
#define inf 1<<30
#define eps 1e-7
#define LD long double
#define LL long long
#define maxn 100000005
int pre[50009] = {};
int rootsearch(int root)//寻找这个数的领导
{
    int son;
    son = root;
    while (root != pre[root])//这个数不是领导，继续往下找
    {
        root = pre[root];
    }
    while (son != root)//路径压缩，把找到的下级数全部指向最高领导
    {
        int temp = pre[son];
        pre[son] = root;
        son = temp;
    }
    return root;//返回最高领导
}
int main()
{
    int n, m, total;
    int Case = 1;
    while (~scanf("%d%d", &n, &m))
    {
        if (n == 0 && m == 0)break;
        memset(pre, 0, sizeof(pre));
        total = n;//n个集合
        for (int i = 1; i <= n; i++)//一开始，自己是自己的领导
        {
            pre[i] = i;
        }
        while (m--)
        {
            int x, y;
            int root1, root2;
            scanf("%d%d", &x, &y);
            root1 = rootsearch(x);
            root2 = rootsearch(y);
            if (root1 != root2)//两个人的最高领导不同
            {
                pre[root1] = root2;//把第一个集合的最高领导改为第二个集合的最高领导
                total--;//集合数减一
            }
        }
        printf("Case %d: %d\n",Case++, total);
    }
    return 0;
}

 

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