並查集的應用：hdu 1213

風穆發表於2016-11-22
原文網址 : https://blog.csdn.net/xingdragon/article/details/53291361
並查集
Problem Description
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









AC程式碼：




# include <stdio.h>

int group[1005];
int depth[1005];


void init()//初始化組別
{
	int i;
	for (i = 0; i < 1005; i ++)
	{
		group[i] = i;
		depth[i] = 1;
	}
}

int find(int e) //查詢根並把同組的都連到根部
{
	if (group[e] == e)
	{
		return e;
	}
	else
	{
		return group[e] = find(group[e]);
	}
}

void unite(int e1, int e2) //連線兩個不同的組
{
	e1 = find(e1);
	e2 = find(e2);

	if (depth[e1] < depth[e2]) //把樹的深度小 的連線到深度大的
	{
		group[e1] = e2;
	}
	else
	{
		group[e2] = e1;
		if (depth[e1] == depth[e2])
		{
			depth[e1] ++;
		}
	}
}


int main(void)
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		int n, m;
		int i;
		int count = 0;	
		init();
		scanf("%d %d", &n, &m);
		for (i = 1; i <= m; i++)
		{
			int head, tail;
			scanf("%d %d", &head, &tail);
			unite(head, tail);
		}
	
		for (i = 1; i <= n; i++)
		{
			if (group[i] == i)
			{
				count++;
			}
		}
		printf("%d\n", count);
	}
	return 0;
}







                                    




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