DHUOJ 2016060703 - Coach（并查集）

最新推荐文章于 2023-04-08 18:06:33 发布

原创最新推荐文章于 2023-04-08 18:06:33 发布 · 298 阅读

0 ·

CC 4.0 BY-SA版权

------题解------ 专栏收录该内容

179 篇文章

订阅专栏

探讨如何通过并查集算法解决ACM训练队的学生分组问题，确保每组三人且互为朋友关系，实现最优分组。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Coach

From: DHUOJ, 2016060703

&lt;a type="button" class="btn btn-warning" href="/solution/submit.html?problemId=5270"&gt;Submit&lt;/a&gt;

Time Limit: 1 s

Description

In the ACM training team, the coach wants to split all students into many groups and each group consists of three students. Let's assume that all students are numbered from 1 to n . The teacher wants the groups to achieve good results, so he wants to hold the following condition: In a group of three students, they are friends. Also it is obvious that each student must be in exactly one team. If A and B are friends, B and C are friends, then A and C are friends.

Input

There are multiple test cases.

For each test case, the first line contains integers n and m (1≤n≤5000,0≤m≤5000) . Then follows m lines, each contains a pair of integers a i ,b i (1≤a i ,b i ≤n) . The pair a i ,b i means that the students a i and b i are friend. It is guaranteed that each pair of a i and b i occurs in the input at most once.

Output

If the required group division plan doesn't exist, print “No”. Otherwise, print “Yes”.

Sample Input

Sample Output

No
Yes

Author: betty

思路：

每三个人组成一组，要求他们三个要有相同的父亲。求能否正好分成若干组，无剩余的人。

经典的并查集问题，因为每个人都会有一个父亲，也就是说，有几种父亲就代表了几个团体，然后就看每个团体里面是否正好是3n个人，若不是，则不能构成。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<stack>
using namespace std;
 
int  pre[5050];  
int t[5050]={0};           
int find(int x)  
{  
    int r=x;  
    while(r!=pre[r])  
        r=pre[r];  
       
    int i=x,j;  
    while(pre[i]!=r)  
    {  
        j=pre[i];  
        pre[i]=r;  
        i=j;  
    }  
    return r;  
}  
   
void merge(int x,int y)  
{  
    int fx=find(x),fy=find(y);  
    if(fx!=fy)  
    {  
        pre[fy]=fx;  
    }  
}   
 
int main()  
{  
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        memset(pre,0,sizeof(pre));
        bool flag=1;
        for(int i=1;i<=n;i++){
            pre[i]=i;
        }
        for(int i=0;i<m;i++){
            int a,b;
            scanf("%d%d",&a,&b);
            merge(a,b);
        }
        memset(t,0,sizeof(t));  
        for(int i=1;i<=n;i++)   
        {  
            t[find(i)]++;
        }  
                  
        for(int i=1;i<=n;i++)   
        {  
//              cout<<i<<" "<<t[find(i)]<<" "<<find(i)<<" "<<endl;
            if(t[find(i)]%3!=0){
                flag=0;
                break;
            }  
        }
        if(flag==1)cout<<"Yes"<<endl;
        else cout<<"No"<<endl;  
    }  
 
    return 0;  
}