HDU - 1213 How Many Tables （并查集）

生日派对桌位最少需求算法

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本博客介绍了一个基于并查集算法的派对桌位分配问题解决方案。问题涉及多个案例，每个案例中，输入N个朋友和M对相互认识的朋友关系，目标是最小化所需桌子数量，确保每桌上的朋友都互相认识。通过并查集进行朋友关系的集合划分，最终输出所需桌子的最小数量。

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
问题链接：http://acm.hdu.edu.cn/showproblem.php?pid=1213
问题简述：有T个案例，每个案例第一行输入N,M，表示有N个朋友，一下有M行。每行输入两个数字表示这两个朋友互相认识。只能让两两认识的朋友坐在一起，求最小的桌子数量
问题分析：并查集水题…求集合数量即可
AC通过的C++语言程序如下：

#include<iostream>
#include <algorithm>
#include<iostream>
#include<string>
#include<stdio.h>
#include <algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
using namespace std;
const int maxn = 1005;
int f[maxn];
int find(int x)
{
	if (f[x] != x)
	{
		f[x] = find(f[x]);
	}
	return f[x];
}
int main()
{
	int n;
	cin >> n;
	for (int i = 0; i < n; i++)
	{
		int fri, line, res = 0;
		cin >> fri >> line;
		for (int i = 0; i < maxn; i++)
		{
			f[i] = i;
		}
		for (int j = 1; j <= line; j++)
		{
			int a, b;
			cin >> a >> b;
			int c = find(a), d = find(b);
			if (c != d)
			{
				f[d] = c;
			}
		}
		for (int j = 1; j <= fri; j++)
		{
			find(j);
			if (find(j) == j)
			{
				res++;
			}
		}
		cout << res << endl;
	}
	return 0;
}