How Many Tables HDU - 1213

原创于 2020-11-27 21:12:50 发布 · 79 阅读

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本文介绍了一种基于并查集的算法，用于解决在生日聚会上如何根据朋友间的熟识关系最少分配餐桌的问题。该算法将熟识的朋友视为同一集合，并通过合并集合的方式减少所需餐桌的数量。

 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

题意：有n个人，每次输入两个人是朋友，不同朋友群体们组成一个个集合，问有多少个集合。（可能语义不太清楚，大家还是看英文吧）
思路：并查集，基础题。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 1e5 + 5;
int v[1010];
double a[maxn];
int ans=0;
int search(int x)
{
	if (a[x] != x)
		search(a[x]);
	else return a[x];
}
int main()
{
	int t;
	cin >> t;
	while (t--)
	{

		int m, n;
		cin >> m >> n;
		for (int i = 1; i <= m; i++)
			a[i] = i;
		for (int i = 1; i <= n; i++)
		{
			int x, y;
			cin >> x >> y;
			
		//		cout << a[search(x)] << endl;
				a[search(x)] = a[search(y)]; 	
			
		}
		int ans = 0;
		for (int i = 1; i <= m; i++)
			if (i == a[i]) ans++;
		cout << ans << endl;         
	}
	return 0 ;
}