poj2002--Squares(n個點求正方形個數)

poj2002--Squares(n個點求正方形個數)

Squares Time Limit: 3500MS Memory Limit: 65536K Total Submissions: 16615 Accepted: 6320

Description

A square is a 4-sided polygon whose sides have equal length and adjacent sides form 90-degree angles. It is also a polygon such that rotating about its centre by 90 degrees gives the same polygon. It is not the only polygon with the latter property, however, as a regular octagon also has this property.

So we all know what a square looks like, but can we find all possible squares that can be formed from a set of stars in a night sky? To make the problem easier, we will assume that the night sky is a 2-dimensional plane, and each star is specified by its x and y coordinates.

Input

The input consists of a number of test cases. Each test case starts with the integer n (1 <= n <= 1000) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (two integers) of each point. You may assume that the points are distinct and the magnitudes of the coordinates are less than 20000. The input is terminated when n = 0.

Output

For each test case, print on a line the number of squares one can form from the given stars.

Sample Input

4
1 0
0 1
1 1
0 0
9
0 0
1 0
2 0
0 2
1 2
2 2
0 1
1 1
2 1
4
-2 5
3 7
0 0
5 2
0

Sample Output

1
6
1

給出n個點，求出有可以組成多少個正方形？

枚舉對角的兩個點，然後求解出其他的兩個點，將這兩個點帶入到n個點中查找，可以hash 或 二分。

已知對角線的點 (x1,y1) (x3,y3) 求出中點( (x1+x3)/2 , (y1+y3)/2 ) ->（X,Y） 用對角線的一個點減去中點得到(x,y)，那麼其他的兩個點就是( x1-Y,y1+X ) （x1+Y,y1-X）

#include 
#include 
#include 
#include 
#include 
using namespace std ;
#define eqs 1e-9
struct node{
    double x , y ;
}p[1100] ;
bool cmp(node a,node b)
{
    return ( a.x < b.x || ( a.x == b.x && a.y < b.y ) ) ;
}
bool judge(double x,double y,int n)
{
    int low = 0 , mid , high = n-1 ;
    while( low <= high )
    {
        mid = (low + high) / 2 ;
        if( fabs(p[mid].x-x) < eqs && fabs(p[mid].y-y) < eqs )
            return true ;
        else if( p[mid].x-x > eqs || ( fabs(p[mid].x-x) < eqs && p[mid].y-y > eqs ) )
            high = mid - 1 ;
        else
            low = mid + 1 ;
    }
    return false ;
}
int main()
{
    int n , i , j , num ;
    double x , y , xx , yy ;
    while( scanf("%d", &n) && n )
    {
        num = 0 ;
        for(i = 0 ; i < n ; i++)
        {
            scanf("%lf %lf", &p[i].x, &p[i].y) ;
        }
        sort(p,p+n,cmp) ;
        for(i = 0 ; i < n ; i++)
        {
            for(j = i+1 ; j < n ; j++)
            {
                if( i == j ) continue ;
                x = (p[i].x+p[j].x)/2 ;
                y = (p[i].y+p[j].y)/2 ;
                xx = p[i].x - x ;
                yy = p[i].y - y ;
                if( judge(x+yy,y-xx,n) && judge(x-yy,y+xx,n) )
                {
                    //printf("(%.1lf,%.1lf) (%.1lf,%.1lf) (%.1lf,%.1lf) (%.1lf,%.1lf)\n", p[i].x, p[i].y , p[j].x, p[j].y,x+yy,y-xx,x-yy,y+xx ) ;
                    num++ ;
                }
            }
        }
        printf("%d\n", num/2) ;
    }
    return 0;
}