hdu2428 Stars

最新推荐文章于 2020-02-17 21:36:23 发布

原创最新推荐文章于 2020-02-17 21:36:23 发布 · 378 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#哈希表

哈希表专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一个利用全哈希解决寻找星空中由星星构成的正方形数量的问题。通过构造哈希表，算法有效地判断了四颗星星是否能形成边平行于坐标轴的正方形，并给出了具体实现细节。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Stars

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1764 Accepted Submission(s): 612

Problem Description

Lucy loves stars very much. There are N (1 <= N <= 1000) stars in the sky. Assume the sky is a flat plane. All of the stars lie on it with a location (x, y), -10000 <= x, y <= 10000. Now, Lucy wants you to tell her how many squares with each of four vertices formed by these stars, and the edges of the squares should parallel to the coordinate axes.

Input

The first line of input is the number of test case.
The first line of each test case contains an integer N. The following N lines each contains two integers x, y, meaning the coordinates of the stars. All the coordinates are different.

Output

For each test case, output the number of squares in a single line.

Sample Input

Sample Output

0
1

全哈希。

#include<cstdio>
#include<cstring>
#include<string.h>

using namespace std;
const int maxn=1005;
const int hashsize=2006;
struct Point
{
	int x;
	int y;
	Point(int a=0,int b=0)
	{
		x=a;
		y=b;
	}
}stars[maxn];

int head[hashsize];
int pre[maxn];

int hash(Point p)
{
	int c=hashsize;
	return ((p.x*20005+p.y)%c+c)%c;
}

int find(int x,int y)
{
	int f=hash(Point(x,y));
	for(int i=head[f];i>=0;i=pre[i])
	{
		if(stars[i].x==x&&stars[i].y==y)
		return 1;
	}
	return 0;
}

int main()
{
	int casen;
	scanf("%d",&casen);
	while(casen--)
	{
		int n;
		scanf("%d",&n);
		memset(head,-1,sizeof(head));
		for(int i=0;i<n;i++)
		{
			scanf("%d%d",&stars[i].x,&stars[i].y);
			int f=hash(stars[i]);
			pre[i]=head[f];
			head[f]=i;
		}
		int sum=0;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(stars[i].x>=stars[j].x)
				continue;
				if(stars[i].x-stars[j].x!=stars[i].y-stars[j].y)
				continue;
				if(!find(stars[i].x,stars[j].y))continue;
				if(!find(stars[j].x,stars[i].y))continue;
				sum++;
			}
		}
		printf("%d\n",sum);
	}
	return 0;
}