HDOJ 1213 How Many Tables（基础并查集）

最新推荐文章于 2017-08-20 21:53:22 发布

AnICoo1

最新推荐文章于 2017-08-20 21:53:22 发布

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分类专栏：并查集

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本文介绍了一个算法问题——如何计算最少需要多少张餐桌来安排一群相互认识的朋友就座，确保不熟悉的人不会坐在同一桌。通过并查集算法解决该问题，并提供了一段完整的AC代码作为实现示例。

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How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17998 Accepted Submission(s): 8856

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


和畅通工程差不多，就是求分成了几部分
ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int pri[100005];
int num[100005];
int find(int x)
{
	int r=x;
	while(r!=pri[r])
	{
		r=pri[r];
	}
	int i=x,j;
	while(i!=r)
	{
		j=pri[i];
		pri[i]=r;
		i=j;
	}
	return r;
}
void connect(int xx,int yy)
{
	int a=find(xx);
	int b=find(yy);
	if(a!=b)
	pri[b]=a;
}
int main()
{
	int t;
	int n,m,i,f,d;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(i=1;i<=n;i++)
		pri[i]=i;
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&f,&d);
			connect(f,d);
		}
		memset(num,0,sizeof(num));
		for(i=1;i<=n;i++)
		num[find(i)]=1;
		int sum=0;
		for(i=1;i<=n;i++)
		if(num[i])
		sum++;
		printf("%d\n",sum);
	}
	return 0;
}