Poj2002--squares (n points to find the number of squares)

Squares

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

Description

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

So we are know what's a square looks like, but can we find all possible squares that can is formed from a set of stars in a Night sky? To make the problem easier, we'll assume that the night sky was a 2-dimensional plane, and each star was specified by its X and Y coordinates. 

Input

The input consists of a number of test cases. Each test case is starts with the integer n (1 <= n <=) indicating the number of points to follow. Each of the next n lines specify the x and y coordinates (both integers) of each point. Assume that the points is distinct and the magnitudes of the coordinates is 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

41 00 11 10 090 01 02 00 21 22 20 11 12 14-2 53 70 05 20

Sample Output

161

Give n points to find out how many squares can be formed?

Enumerate the two points diagonally, then solve the other two points, take the two points into the N points to find, you can hash or two points.

The point of the known diagonal (x1,y1) (X3,Y3) is calculated by the midpoint ((X1+X3)/2, (Y1+Y3)/2) (x, y) minus the midpoint with one point on the diagonal (x, y), then the other two points are (x1-y,y1+x) (x1+y,y1-x)

#include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <vector >using namespace std; #define EQS 1e-9struct 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, &AMP;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;}

Poj2002--squares (n points to find the number of squares)