HDU-1213 How Many Tables

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这篇文章介绍了一个关于计算Ignatius生日聚会所需最少餐桌问题的算法，根据朋友之间的相互认识关系来确定。输入为朋友数量和认识关系的数量，输出为最少餐桌数。核心在于合并已知熟识的朋友群，避免陌生人共桌。

How Many Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 73145 Accepted Submission(s): 36926

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

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#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int p[1010];

int find(int x)
{
	if(p[x]!=x)
	return p[x]=find(p[x]);
	return x;
}
int merge(int x,int y)
{
    int fx=find(x);
    int fy=find(y);
    if(p[fx]!=p[fy])
    {
        p[fx]=fy;
        return 1;
	}
	return 0;
}
int main()
{
	int t;
	int m,n;
	int x,y;
	int ans;
	
	cin>>t;
	while(t--)
	{
		cin>>n>>m;
		ans=n;
		for(int i=1;i<=n;i++)p[i]=i;
		for(int i=1;i<=m;i++)
		{
			cin>>x>>y;
			if(merge(x,y))
			ans-=1;
		}
		cout<<ans<<endl;
	}
	return 0;//
}