hdu 1213 How Many Tables

最新推荐文章于 2025-08-13 23:15:00 发布

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最新推荐文章于 2025-08-13 23:15:00 发布

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本文链接：https://blog.youkuaiyun.com/yyyy2013_8_3/article/details/9901711

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本文探讨如何运用并查集算法解决派对中不同朋友间的关系，通过了解朋友之间的相互认识情况，计算最少需要多少张桌子来确保每个人都能够与认识的人相邻就座。通过输入不同数量的朋友及其相互认识的关系，算法输出所需最少桌次。

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How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9359 Accepted Submission(s): 4619

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

题意：朋友可以和朋友或朋友的朋友坐在一起，不会和陌生人坐在一起，求需要几张桌子！（并查集）

AC代码

#include <iostream>
using namespace std;
int father[1111];
int find(int x)
{
	if(father[x]!=x)
	father[x]=find(father[x]);
	return father[x];
}//查找
void Union(int a,int b)
{
    a=find(a);
    b=find(b);
	if(a!=b)
		father[a]=b;
}//并
int main()
{
	int n,m,i,a,b,flag,t;
	while (scanf("%d",&t)!=EOF)
	{
	   while(t--)
	   {
		scanf("%d",&n);
		for (i=1;i<=n;i++)
		    father[i]=i;
		scanf("%d",&m);
		while(m--)
		{
			scanf("%d%d",&a,&b);
			Union(a,b);
		}
		flag=0;
		for (i=1;i<=n;i++)
		{
			if(father[i]==i)
				flag++;
		}
		printf("%d\n",flag);//输出剩余的二叉树数
	   }	
	}
	return 0;
}

谢谢阅读！