HDU 1856 并查集从这个题入手比较容易懂

最新推荐文章于 2021-05-21 17:54:02 发布

转载最新推荐文章于 2021-05-21 17:54:02 发布 · 480 阅读

本文介绍了一个寻找最大朋友圈的算法问题，通过并查集的方法解决如何从大量男孩中找出最大组直接或间接朋友群体的问题。该算法使用了初始化、查找、合并及计数等步骤来高效解决问题。

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More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)

Total Submission(s): 24557 Accepted Submission(s): 8806

Problem Description

Mr Wang wants some boys to help him with a project. Because the project is rather complex,the more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

Input

The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

Output

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

Sample Input

Sample Output

4
2

Hint
A and B are friends(direct or indirect), B and C are friends(direct or indirect), 
then A and C are also friends(indirect).

 In the first sample {1,2,5,6} is the result.
In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.

题目大意是，找到最大是的朋友圈，

#include <iostream>
using namespace std;

const int MAXN = 10000005;

struct Node 
{
	int parent;//保存父节点
	int rank;//以此小男孩为父节点的人数
};
Node boy[MAXN];

void init(void)
{
	int i;
	for (i = 0; i < MAXN; i++)
	{
		boy[i].parent = i;
		boy[i].rank = 1;
	}
}

int find(int a)  //寻找 a 的父节点
{
	int t;
	t = boy[a].parent;
	if (t != a)    <span style="font-family: Arial, Helvetica, sans-serif;">//直到找到父节点为止</span>
	{
		boy[a].parent = find(t);   
	}
	return boy[a].parent;
}

void Union(int a, int b)
{
	int fa, fb;
	fa = find(a);
	fb = find(b);
	if (fa != fb)  
	{
		boy[fa].parent = fb;     //fa fb 的父节点不同，它 fa  归属于 fb  
		boy[fb].rank += boy[fa].rank;    //统计以 fb 为父节点的数目
	}
}

int Count()
{
	int i;
	int cnt = 0;
	for (i = 0; i < MAXN; i++)
	{
		if (boy[i].parent == i)    //已经归属别人的就不用再管了
		{
			if (boy[i].rank > cnt)
				cnt = boy[i].rank;
		}
	}
	return cnt;
}

int main(void)
{
	int tcase;
	int a, b;

	while (scanf("%d", &tcase) != EOF)
	{
		init();
		while (tcase--)
		{
			scanf("%d%d", &a, &b);
			Union(a, b);
		}
		printf("%d\n", Count());
	}

	return 0;
}