POJ 2002 Squares (二分)

Squares

Time Limit : 7000/3500ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 12   Accepted Submission(s) : 11

Problem 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

PKU

 

 

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

struct Point{
    int x,y;
    Point(){}
    Point(int a,int b):x(a),y(b){}
}p[1010];

int n;

int cmp(Point a,Point b){
    if(a.x==b.x)
        return a.y<b.y;
    return a.x<b.x;
}

int BinSearch(Point tmp){
    int l=0,r=n,mid;
    while(l<=r){
        mid=(l+r)>>1;
        if(p[mid].x==tmp.x && p[mid].y==tmp.y)
            return 1;
        else if(cmp(p[mid],tmp))
            l=mid+1;
        else
            r=mid-1;
    }
    return 0;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d",&n) && n){
        for(int i=0;i<n;i++)
            scanf("%d%d",&p[i].x,&p[i].y);
        sort(p,p+n,cmp);
        int ans=0;
        int x3,y3,x4,y4;
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++){
                x3=p[j].x-(p[j].y-p[i].y);      //给定正方形一条边的两点坐标求另外两点的坐标，自己画图算算
                y3=p[j].y+(p[j].x-p[i].x);
                if(!BinSearch(Point(x3,y3)))
                    continue;
                x4=p[i].x-(p[j].y-p[i].y);
                y4=p[i].y+(p[j].x-p[i].x);
                if(!BinSearch(Point(x4,y4)))
                    continue;
                ans++;
                /*
                printf("-------------\n");
                printf("%d %d %d %d %d %d %d %d\n",p[i].x,p[i].y,p[j].x,p[j].y,x3,y3,x4,y4);
                printf("-------------\n");
                */
            }
        printf("%d\n",ans/2);
    }
    return 0;
}