HDU 1213 how many tables

最新推荐文章于 2020-07-24 00:09:49 发布

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#并查集 #c语言 #算法

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本文介绍了一个用于解决生日派对中桌位安排问题的算法。该算法基于并查集思想，确保互相认识的朋友可以坐在同一张桌子上，同时尽可能减少所需桌子的数量。文章通过示例解释了算法的工作原理，并提供了完整的C++代码实现。

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#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
#include<queue>

using namespace std;

int visitor[1002];
int already[1002];

void chu(int x)
{
	for (int i = 1; i <= x; i++)
	{
		visitor[i] = i;
	}
}

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

bool friendfind(int x)
{
	for (int i = 0; already[i] != 0; i++)
	{
		if (already[i] == x)
		{
			return false;
		}
	}
	return true;
}

int main()
{
	int n;
	cin >> n;
	while (n--)
	{
		
		int visitor_num, guan;
		cin >> visitor_num >> guan;
		int v1, v2;
		chu(visitor_num);
		memset(already, 0, sizeof(already));
		for (int i = 0; i < guan; i++)
		{
			cin >> v1 >> v2;
			visitor[find(v1)] = visitor[find(v2)];
		}
		
		int k = 0;
		for (int i = 1; i <= visitor_num; i++)
		{
			int temp = find(visitor[i]);
			if (friendfind(temp))
			{
				already[k++] = temp;
			}
		}
		cout << k<<'\n';
		
	}
	return 0;
}

//by swust_y_p