poj 2524 Ubiquitous Religions [并查集]

最新推荐文章于 2022-12-25 14:52:12 发布

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最新推荐文章于 2022-12-25 14:52:12 发布

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分类专栏：并查集

本文链接：https://blog.youkuaiyun.com/jchangbrave/article/details/52014097

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本文介绍了一种使用并查集数据结构来解决大学校园内学生宗教信仰数量估算的问题。通过询问特定学生对的宗教信仰是否相同，利用并查集进行合并操作，最终确定校园内的最大不同宗教信仰数量。

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Ubiquitous Religions

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 31544		Accepted: 15284

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

Sample Output

Case 1: 1
Case 2: 7

Hint

Huge input, scanf is recommended.

题意：计算n个学生中有多少个不同的宗教信仰。共有m行操作，每一个操作表示这两个学生有共同的宗教信仰。

思路：这个是很明显的并查集的题，每一行的两个学生有相同的宗教信仰，我们可以将有共同宗教信仰的学生合并到一个集合里面，有n名学生，所以说学生中最多的宗教数不超过n，每一次合并就可以减去1个，遍历m行操作后，剩下的宗教数即为所求。

这题提示是要用scanf的，cin处理不好可能会TLE。

写了几道并查集的题，感觉并查集的套路性可以记一下。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;

const int Max=50005;
int p[Max];
//int num[Max];
int rank_s[Max];
int ans;

void make_set(int x)
{
    p[x]=x;
    //num[x]=1;
    rank_s[x]=0;
}

int find_set(int x)
{
    int r=x;
    while(p[r]!=r)
    {
        r=p[r];
    }
    return r;
}

void union_set(int x,int y)
{
    int fx=find_set(x);
    int fy=find_set(y);
    if(fx==fy) return ;
    ans--;
    if(rank_s[fx]>rank_s[fy])
    {
        p[fy]=fx;
       // num[fx]+=rank_s[fy];
    }
    else
    {
        p[fx]=fy;
        //num[fy]+=num[fx];
        if(fx==fy) rank_s[fy]++;
    }
}

int main()
{
    int n,m,cas=0;
    while(scanf("%d%d",&n,&m)&&n&&m)
    {
        for(int i=1;i<=Max;i++) make_set(i);
        ans=n;
        //cout<<ans<<endl;
        while(m--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            union_set(a,b);
        }
        printf("Case %d: %d\n",++cas,ans);
    }
    return 0;
}