HDU 1213（裸并查集）（无变形）

Posted 2020-11-17 yinbiao

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=1213

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 41546    Accepted Submission(s): 20798

Problem Description

Today is Ignatius‘ birthday. He invites a lot of friends. Now it‘s dinner time. Ignatius wants to know how many tables he needs at least. You have to notice that not all the friends know each other, and all the friends do not want to stay with strangers.

One important rule for this problem is that if I tell you A knows B, and B knows C, that means A, B, C know each other, so they can stay in one table.

For example: If I tell you A knows B, B knows C, and D knows E, so A, B, C can stay in one table, and D, E have to stay in the other one. So Ignatius needs 2 tables at least.

 

 

Input

The input starts with an integer T(1<=T<=25) which indicate the number of test cases. Then T test cases follow. Each test case starts with two integers N and M(1<=N,M<=1000). N indicates the number of friends, the friends are marked from 1 to N. Then M lines follow. Each line consists of two integers A and B(A!=B), that means friend A and friend B know each other. There will be a blank line between two cases.

 

 

Output

For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

 

 

Sample Input

2 5 3 1 2 2 3 4 5 5 1 2 5

 

 

Sample Output

2 4

 

 

Author

Ignatius.L

 

 

Source

杭电ACM省赛集训队选拔赛之热身赛

 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 T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        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("%d
",ans);
    }
    return 0;
}