Poj2002: squares

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 center 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-dimen1_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

Enumerate two adjacent points, find the other two points, and check whether these two points exist.

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define up(i,x,y) for(i=x;i<=y;i++)#define down(i,x,y) for(i=x;i<=y;i++)#define mem(a,b) memset(a,b,sizeof(a))#define w(a) while(a)const int mod=20007;int n,next[20007],head[20007],m,ans;struct node{    int x,y;} a[2222];int cmp(node a,node b){    if(a.x==b.x)        return a.y<b.y;    return a.x<b.x;}void insert(int i){    int key=(a[i].x*a[i].x+a[i].y*a[i].y)%mod;    next[m]=head[key];    a[m].x=a[i].x;    a[m].y=a[i].y;    head[key]=m++;}int find(int x,int y){    int key=(x*x+y*y)%mod;    for(int i=head[key];i!=-1;i=next[i])    if(a[i].x==x&&a[i].y==y)    return i;    return -1;}int main(){    int i,j;    w((scanf("%d",&n),n))    {        mem(head,-1);        mem(next,0);        m=1005;        ans=0;        up(i,0,n-1)        {            scanf("%d%d",&a[i].x,&a[i].y);            insert(i);        }        sort(a,a+n,cmp);        up(i,0,n-1)        {            up(j,i+1,n-1)            {                int x1,y1,x2,y2;                x1=a[i].x-a[j].y+a[i].y;                y1=a[i].y+a[j].x-a[i].x;                if(find(x1,y1)==-1)                continue;                x2=a[j].x-a[j].y+a[i].y;                y2=a[j].y+a[j].x-a[i].x;                if(find(x2,y2)==-1)                continue;                ans++;            }        }        printf("%d\n",ans/2);    }    return 0;}