PKU 3432 Count Squares

Count Squares

Time Limit: 3000MS
Memory Limit: 65536KTotal Submissions: 1331
Accepted: 552

Description

Given a set of points with integer coordinates x_i, y_i, i = 1...N, your program must find all the squares having each of four vertices in one of these points.

Input

Input file contains integer N followed by N pairs of integers x_i y_i.

Constraints

-10⁴ ≤ x_i, y_i ≤ 10⁴, 1 ≤ N ≤ 2000. All points in the input are different.

Output

Output file must contain a single integer — number of squares found.

Sample Input

Sample input 1
4 0 0 4 3 -3 4 1 7
Sample input 2
9
1 1  1 2  1 3  
2 1  2 2  2 3  
3 1  3 2  3 3

Sample Output

Sample output 1
1
Sample output 2
6

Hint

Bold texts appearing in the sample sections are informative and do not form part of the actual data.

Source

Northeastern Europe 2005, Far-Eastern Subregion

题目意思很简单，给一群不同的点，求这些点最多组成了几个正方形。

每此枚举2个点，作为正方行的一条边，然后顺时针/逆时针旋转90度(利用相对坐标来计算)，得到另外2个点的坐标，查询HASH看这2个点如果都存在那么cnt++;
最后cnt/4,因为一个正方形被算过4次

1. for(i=0;i<n;i++)
2.             for(j=0;j<n;j++)
3.             {
4.                 if(i==j)continue;
5.                                 //下面是重点-坐标变换
6.                 TMP.x=P[j].y-P[i].y+P[j].x;
7.                 TMP.y=P[i].x-P[j].x+P[j].y;
8.                 if(isin(TMP))//isin查询HASH表中点是否存在
9.                 {
10.                                         //坐标变换，得到另一个点
11.                     TMP.x+=(P[i].x-P[j].x);
12.                     TMP.y+=(P[i].y-P[j].y);
13.                     if(isin(TMP))
14.                         cnt++;
15.                 }
16.             }
17.         printf("%d/n",cnt/4);