二分。。。。

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 14530		Accepted: 5488

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

 

枚举两个点计算另外俩个，二分查找。。。。。

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

struct node

{

    int x,y;

}p[1100],p1,p2;

int n,cnt;

bool cmp(node a,node b)

{

    if(a.x!=b.x) return a.x<b.x;

    return a.y<b.y;

}

bool findp(node a)

{

    int s=0,t=n-1,m;

    int X=a.x,Y=a.y;

    while(s<=t)

    {

        m=(s+t)/2;

        if(p

.x==X&&p

.y==Y) 

return 

true;

        

if(p

.x<X||(p

.x==X&&p

.y<Y)) s=m+

1;

        

else t=m-

1;

    }

    

return 

false;

}

int main()

{

    

while(scanf(

"%d",&n)!=EOF&&n)

    {

        cnt=

0;

        

for(

int i=

0;i<n;i++)

        {

            scanf(

"%d%d",&p

.x,&p.y);

        }

        sort(p,p+n,cmp);

        for(int i=0;i<n;i++)

        {

            for(int j=i+1;j<n;j++)

            {

                p1.x=p.x+p[j].y-p.y; p1.y=p.y+p.x-p[j].x;

                p2.x=p[j].x+p[j].y-p.y; p2.y=p[j].y+p.x-p[j].x;

                if(!findp(p1)) continue;

                if(findp(p2)) cnt++;

            }

        }

        printf("%d\n",cnt/2);

    }

    return 0;

}

* This source code was highlighted by . ( style: Codeblocks )