D - How Many Tables 【hdu 1213 并查集】

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本文介绍了一个经典的图论问题——如何根据朋友间的相识关系最少地安排餐桌数。通过并查集算法解决该问题，并提供了完整的代码实现。

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题目链接

How Many Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 32484 Accepted Submission(s): 16206

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

入门题目，与畅通工程有点像。。。。

#include<cstdio>
int pre[1010];
int find(int a)//find函数查找他的领导 
{
	int leader=a;
	while(pre[leader]!=leader)//领导不是自己，找自己的上一级领导 
		leader=pre[leader]; 
	int t,b=a;
	while(b!=leader)
	{
		t=pre[a];
		pre[a]=leader;
		b=t;
	}
	return leader;
}
int main()
{
	int t,n,m,point1,point2,all,lead1,lead2;//point表示城镇编号 
	while(~scanf("%d",&t))
	{
		while(t--){
			scanf("%d%d",&n,&m);
			all=n;//n个人需要n个桌子 
			for(int i=1;i<=n;i++){
				pre[i]=i;//初始化为自己的领导是自己 
			}
			for(int j=0;j<m;j++){
				scanf("%d%d",&point1,&point2);
				lead1=find(point1);
				lead2=find(point2);
				if(lead1!=lead2){
					pre[lead2]=lead1;
					all--;//根据题目给出的条件每认识一个人总数减一 
				}
			}
			printf("%d\n",all);
		}
	}
	return 0;
}