[并查集]PJOJ 2524 圣主在上

最新推荐文章于 2021-04-15 03:51:16 发布

原创最新推荐文章于 2021-04-15 03:51:16 发布 · 307 阅读

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本文介绍了一个使用并查集算法解决的问题：通过询问学生之间的宗教信仰相似性来估算大学校园内不同宗教信仰的最大数量。文章提供了完整的C++代码实现，并解释了如何通过并查集来统计不同子树的数量，以此推算出可能存在的最大宗教种类数。

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.
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

分析
裸并查集，统计不同子树个数

#include <iostream>
#include <cstdio>
#define rep(i,a,b) for (i=a;i<=b;i++)
using namespace std;
int n,m;
int f[50001];
int b[50001];
int ans;
int i,count;
int getf(int x)
{
    if (f[x]==x) return x;
    int root=getf(f[x]);
    f[x]=root;
    return root;
}
void add(int x,int y)
{
    int i=getf(x),j=getf(y);
    f[i]=j;
}
int main()
{
    while (1)
    {
        scanf("%d%d",&n,&m);
        count++;
        if (n==m&&n==0) break;
        rep(i,1,n)
        {
            b[i]=0;
            f[i]=i;
        }
        ans=0;
        rep(i,1,m)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            add(a,b);
        }
        rep(i,1,n)
        {
            int father=getf(i);
            if (!b[father])
            {
                b[father]=1;
                ans++;
            }
        }
        printf("Case %d: %d\n",count,ans);
    }
}