HDU 1213 How Many Tables

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探讨如何通过算法计算在确保每位朋友都不会与陌生人同桌的前提下，Ignatius为生日派对所需准备的最少餐桌数量。该问题通过并查集算法解决，确保了已知相互认识的朋友能够坐在同一桌。

How Many Tables

https://vjudge.net/problem/HDU-1213
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
C++

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 1000 + 10;
int fa[MAXN];
int n, m;

void init()
{
	for(int i = 1; i <= n; i ++)
		fa[i] = i;
}

int find(int x)
{
	if(fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}

void merge(int x, int y)
{
	int tx = find(x);
	int ty = find(y);
	if(tx != ty)
		fa[tx] = ty;
}

int main()
{
	int T;
	scanf("%d", &T);
	while(T --)
	{
		scanf("%d%d", &n, &m);
		init();
		int a, b;
		for(int i = 0; i < m; i ++)
		{
			scanf("%d%d", &a, &b);
			merge(a, b);
		}
		int ans = 0;
		for(int i = 1; i <= n; i ++)
			if(fa[i] == i) ans ++;
		printf("%d\n", ans);
	}
	return 0;
}