POJ2524 Ubiquitous Religions (并查集)

最新推荐文章于 2022-12-25 14:52:12 发布

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本文介绍了一种算法，用于估算大学校园内不同宗教信仰的最大可能数量。通过询问学生两两之间的宗教信仰是否相同，并利用并查集算法进行处理，可以在不直接获取每个学生宗教信仰的情况下，推算出信仰种类的上限。

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

Sample Output

Case 1: 1
Case 2: 7

Hint
Huge input, scanf is recommended.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
#define MAXN 50050
int s=0;
int pre[30030];
int find(int x)
{
    int r = x;      //用r来传递信息
    while(pre[r] != r)  //如果r的上级bushir自身
        r = pre[r];     //r就接着找上级，直到找到老大
    return r;           //告诉你x的老大是谁
}

void join(int x, int y)     //x，y成为朋友，于是他们两个的联盟合并了
{
    int fx = find(x);       //找x的老大
    int fy = find(y);       //找y的老大
    if(fx != fy){
        pre[fx] = fy;       //x的老大成了y老大的小弟
    }
}

int main()
{
    int n, m, i, j;
    while(cin >> n >> m&&( n||m))
    {
        s++;
        int sum = 0;
        for(j=1; j<=n; j++){
            pre[j] = j;
        }
        while(m--)
        {
            int a, b;
            cin >> a >> b;
            join(a, b);
        }
        for(j=1; j<=n; j++){
            if(pre[j]==j)
                sum++;
        }
        cout << "Case " << s << ": " <<sum << endl;
    }
    return 0;
}