Squares

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

已知： (x1,y1) (x2,y2)

则： x3=x1+(y1-y2) y3= y1-(x1-x2)

x4=x2+(y1-y2) y4= y2-(x1-x2)

或

x3=x1-(y1-y2) y3= y1+(x1-x2)

x4=x2-(y1-y2) y4= y2+(x1-x2)

代码一
#include <cmath>
#include <queue>
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
 
typedef struct point
{
    int x;
    int y;
}PO;
 
bool cmp(PO x,PO y)
{
    if(x.x == y.x) return x.y < y.y;
    else return x.x < y.x;
}
 
void trans(PO i,PO j,PO &C,PO &D,int f)
{
    int tx = i.x - j.x;
    int ty = i.y - j.y;
    C.x = i.x + f * ty;
    C.y = i.y - f * tx;
    D.x = j.x + f * ty;
    D.y = j.y - f * tx;
}
 
bool find(PO P[],int n,PO C)
{
    int l,r;
    l = 0,r = n - 1;
    while(l <= r)
    {
        int mid = (l + r) / 2;
        if(P[mid].x == C.x)
        {
            if(P[mid].y == C.y) return true;
            if(P[mid].y > C.y)
            {
                r = mid - 1;
            }
            if(P[mid].y < C.y)
            {
                l = mid + 1;
            }
        }
        if(P[mid].x < C.x)
        {
            l = mid + 1;
        }
        if(P[mid].x > C.x)
        {
            r = mid - 1;
        }
    }
    return false;
}
 
int main()
{
    int n;
    while(scanf("%d",&n) && n)
    {
        PO P[1000];
        int ans = 0;
        for(int i = 0;i < n;i++)
        {
            scanf("%d%d",&P[i].x,&P[i].y);
        }
        sort(P,P + n,cmp);
        for(int i = 0;i < n;i++)
        {
            for(int j = i + 1;j < n;j++)
            {
                PO C,D;
                trans(P[i],P[j],C,D,1);
                if(find(P,n,C) && find(P,n,D)) ans++;
                trans(P[i],P[j],C,D,-1);
                if(find(P,n,C) && find(P,n,D)) ans++;
            }
        }
        printf("%d\n",ans / 4);
    }
}

代码二
#include <iostream>
#include <stdio.h>
#include <set>
#include <cmath>
#include <algorithm>
#define e 1e-9
using namespace std;
struct node
{
    double x;
    double y;
}p[1010];
bool judege(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)<e&&fabs(p[mid].y-y)<e)
            return 1;
        else if(p[mid].x-x>e||(fabs(p[mid].x-x)<e&&p[mid].y-y>e))
            high=mid-1;
        else
            low=mid+1;
    }
    return 0;
}
bool cmp(node a,node b)
{
    return(a.x<b.x||(a.x==b.x&&a.y<b.y));
}
int main()
{
    int n;
    while(cin>>n)
    {
        if(n==0)
            break;
        for(int i=0;i<n;i++)
        {
            scanf("%lf %lf",&p[i].x,&p[i].y);
        }
        sort(p,p+n,cmp);
        int ans=0;
        double xx,yy,x,y;
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
        {
            if(i==j)
                continue;
            xx=(p[i].x+p[j].x)/2;
            yy=(p[i].y+p[j].y)/2;
            x=p[i].x-xx;
            y=p[i].y-yy;
            if(judege(xx-y,yy+x,n)&&judege(xx+y,yy-x,n))
                ans++;
        }
        cout<<ans/2<<endl;
    }
    return 0;
}