How Many Tables

最新推荐文章于 2024-11-10 21:49:48 发布

csu__zhouyi

最新推荐文章于 2024-11-10 21:49:48 发布

阅读量320

点赞数

CC 4.0 BY-SA版权

分类专栏：并查集文章标签：并查集 ACM c++

本文链接：https://blog.youkuaiyun.com/csu__zhouyi/article/details/38108529

并查集专栏收录该内容

2 篇文章

订阅专栏

本文介绍了一道关于朋友分组的经典问题，并通过并查集算法高效解决了该问题。问题要求将相互认识的朋友分配到同一张桌子就餐，且尽量减少桌子的数量。文章提供了完整的AC代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目（hdu1213）：

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

Sample Output

2
4

很显然这道水题用并查集可以轻松解决。

以下是ac的代码：

#include<iostream>
#include<cstring>
using namespace std;
int fa[1005];
int findfa(int x)
{
    while(fa[x]!=x)
        x=fa[x];
    return x;
}

int main()
{
    int t;
    cin>>t;
    for(int i=0;i<t;i++)
    {
        int n,m;
        memset(fa,0,sizeof(fa));
        cin>>n>>m;
        for(int j=0;j<m;j++)
        {

            int x,y;
            cin>>x>>y;
            if(fa[x]==0 && fa[y]==0)
            {
                fa[x]=x;fa[y]=x;
                n--;
            }
            else if(fa[x]==0)
            {
                fa[x]=findfa(y);n--;
            }
            else if(fa[y]==0)
            {
                fa[y]=findfa(x);n--;
            }
            else if(findfa(x)!=findfa(y))
            {
                fa[findfa(x)]=findfa(y);
                n--;
            }

        }
 cout<<n<<endl;
    }
    return 0;
}