并查集 HDU__1213 How Many Tables

How Many Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40259    Accepted Submission(s): 20106

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

互相认识的人坐在一张桌子上，问一共需要多少张桌子，（不考虑桌子最多做多少人）

第一行 t

接下来一行n和m

n是人数，下面m行代表关系

并查集的入门题目

//#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
const int MAXN=1e3+3;
int pre[MAXN];
int t,n,m;
void init(int n)//初始化 
{
	for(int i=1;i<=n;i++)
		pre[i]=i;
} 
int find(int x)//找根 
{
	return pre[x]==x?x:pre[x]=find(pre[x]);
}
void join(int x,int y)//连接 
{
	x=find(x);
	y=find(y);
	if(x!=y)	pre[x]=y;
}
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d %d",&n,&m);
		init(n);
		
		for(int i=0;i<m;i++)
		{
			int x,y;
			scanf("%d %d",&x,&y);
			join(x,y);
		}
		int ans=0;
		for(int i=1;i<=n;i++)
			if(pre[i]==i)	ans++;
		cout<<ans<<endl;
	}
	return 0;
}