poj-2002 Squares

Squares

Time Limit: 3500MS Memory Limit: 65536KTotal Submissions: 11102 Accepted: 4021

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

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

Sample Output

161

#include<iostream>#include<cmath>int n,i,j,ans;using namespace std;struct set{int x,y;};set point[1005];set c,d;int cmp(const void *a,const void *b){if(((set *)a)->x==((set *)b)->x)return ((set *)a)->y-((set *)b)->y;return ((set *)a)->x-((set *)b)->x;}int fab(int a){return a>0?a:-a;}bool search(set p,int n){int max=n,min=1,mid;while(max>=min){mid=(max+min)/2;if(cmp(&point[mid],&p)>0){max=mid-1;}else if(cmp(&point[mid],&p)<0){min=mid+1;}elsereturn true;}return false;}int main(){while(~scanf("%d",&n)){ans=0;if(n==0)break;for(i=1;i<=n;i++)scanf("%d%d",&point[i].x,&point[i].y);qsort(point+1,n,sizeof(set),cmp);for(i=1;i<=n;i++){for(j=i+1;j<=n;j++){int sign=(point[i].x-point[j].x)*(point[i].y-point[j].y);c.x=point[i].x+fab(point[j].y-point[i].y);c.y=point[i].y+fab(point[j].x-point[i].x);d.x=point[j].x+fab(point[j].y-point[i].y);d.y=point[j].y+fab(point[j].x-point[i].x);if(sign<0){if(search(c,n) && search(d,n))ans++;}else if(sign==0 && point[i].x==point[j].x){if(search(c,n) && search(d,n))ans++;}}}printf("%d\n",ans);}return 0;}