Squares

Squares
Time Limit: 3500MS Memory Limit: 65536K
Total Submissions: 17462 Accepted: 6634

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

Source
Rocky Mountain 2004
在推正方形顶点时,多亏了金巨巨,金巨巨就是给力

#include <map>#include <list>#include <cmath>#include <queue>#include <stack>#include <string>#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>using namespace std;#define eps 1e-9#define LL long long#define PI acos(-1.0)#define INF 0x3f3f3f3f#define CRR fclose(stdin)#define CWW fclose(stdout)#define RR freopen("input.txt","r",stdin)#define WW freopen("output.txt","w",stdout)const int MAX = 1010;struct node{    int x;    int y ;}Point[MAX];bool cmp(node b,node c){    if(b.x<c.x||(b.x==c.x&&b.y<c.y))    {        return true;    }    return false;}bool Look(int low,int high,int x,int y)//二分查找{    int i=low,j=high;    while(i<=j)    {        int mid=(i+j)/2;        if(Point[mid].x==x&&Point[mid].y==y)        {            return true;        }        if(Point[mid].x<x)        {            i=mid+1;        }        else if(Point[mid].x>x)        {            j=mid-1;        }        else if(Point[mid].x==x)        {            if(Point[mid].y<y)            {                i=mid+1;            }            else if(Point[mid].y>y)            {                j=mid-1;            }        }    }    return false;}int main(){    int n;    while(scanf("%d",&n),n)    {        for(int i=0;i<n;i++)        {            scanf("%d %d",&Point[i].x,&Point[i].y);        }        sort(Point,Point+n,cmp);        int sum=0;        for(int i=0;i<n;i++)        {            for(int j=i+1;j<n;j++)            {                int ans=(Point[i].x+Point[j].x-Point[j].y+Point[i].y);//计算正方形的其余的两个顶点                int ant=(Point[i].y+Point[j].y+Point[j].x-Point[i].x);                bool flag=false;                if(ans%2==0&&ant%2==0)                {                    flag=Look(0,n-1,ans/2,ant/2);                }                if(!flag)                {                    continue;                }                flag=false;                ans=(Point[i].x+Point[j].x+Point[j].y-Point[i].y);                ant=(Point[i].y+Point[j].y-Point[j].x+Point[i].x);                if(ans%2==0&&ant%2==0)                {                    flag=Look(0,n-1,ans/2,ant/2);                }                if(flag)                {                    sum++;                }            }        }        printf("%d\n",sum/2);//对于每个正方形都会查找到两次,所以除二    }    return 0;}