POJ 2524 独一无二的宗教（裸并查集）

发布时间：2020-12-14 04:47:55 所属栏目：大数据来源：网络整理

导读：题目链接： http://poj.org/problem?id=2524 Ubiquitous Religions Time Limit: 5000MS ? Memory Limit: 65536K Total Submissions: 39369 ? Accepted: 18782 Description There are so many different religions in the world today that it is difficult t

题目链接：

http://poj.org/problem?id=2524

Ubiquitous Religions

Time Limit: 5000MS	?	Memory Limit: 65536K
Total Submissions: 39369	?	Accepted: 18782

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.

分析：

裸的并查集（我觉得叫查并集更好）

代码如下：

#include<stdio.h>
#include<iostream>
using namespace std;
#define max_v 50005
int pa[max_v];//pa[x] 表示x的父节点
int rk[max_v];//rk[x] 表示以x为根结点的树的高度
int n,ans;
void make_set(int x)
{
    pa[x]=x;
    rk[x]=0;//一开始每个节点的父节点都是自己
}
int find_set(int x)//带路径压缩的查找
{
    if(x!=pa[x])
        pa[x]=find_set(pa[x]);
    return pa[x];
}
void union_set(int x,int y)
{
    x=find_set(x);//找到x的根结点
    y=find_set(y);
    if(x==y)//根结点相同 同一棵树
        return ;
    ans--;
    if(rk[x]>rk[y])
    {
        pa[y]=x;
    }else
    {
        pa[x]=y;
        if(rk[x]==rk[y])
            rk[y]++;
    }
}
int main()
{
    int n,m,j=0;
    while(~scanf("%d %d",&n,&m))
    {
        if(m+n==0)
            break;
        for(int i=1;i<=n;i++)
        {
            make_set(i);
        }
        ans=n;
        for(int i=0;i<m;i++)
        {
            int x,y;
            scanf("%d %d",&x,&y);
            union_set(x,y);
        }
        printf("Case %d: %dn",++j,ans);
    }
    return 0;
}

（编辑：李大同）

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