HDU - 1213-day10G

问题来源

HDU - 1213

问题描述

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.

输入

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.

输出

For each test case, just output how many tables Ignatius needs at least. Do NOT print any blanks.

例子

input

2
5 3
1 2
2 3
4 5

5 1
2 5

output

2
4

解题思路

并查集，
输入谁和谁认识。
a认识b，b认识c，等同于a,b,c都认识。
看看有几堆人。

AC的代码

#include <iostream>
#include<cstdio>
using namespace std;
int i[1005], s1, s2, b1, b2;

int find(int j)
{
	int k = j;
	while (i[j] != j)
	{
		j = i[i[j]];
	}
	return j;

}
int main()
{

	int j, k, total,t;
	cin >> t;
	while(t--)
	{
		scanf_s("%d%d", &j, &k);
		total = 0;
		for (int c = 0; c < 1000; c++)
		{
			i[c] = c;
		}
		for (int k1 = 0; k1 < k; k1++)
		{

			scanf_s("%d%d", &s1, &s2);
			b1 = find(s1);
			b2 = find(s2);
			if (b1 != b2)
			{
				i[b1] = b2;
			}
		}

		for (int l1 = 1; l1 <= j; l1++)
		{
			if (i[l1] == l1)
				total++;
		}


		cout << total << endl;
	}
}