poj 2002 二分法的强大应用

        自认为是一道很有难度的题，，同时也再一次体会到了二分的强大，，，，题目：

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

ac代码：

#include <iostream>
#include <algorithm>
#include <string.h>
#include <cstdio>
using namespace std;
struct point
{
  int x,y;
}aa[1005];
int n;
bool cmp(point a,point b)
{
  if(a.x==b.x)
	  return a.y>b.y;
  return a.x<b.x;
}
bool find(int xx,int yy)
{
  int left=0,right=n-1,mid;
  while(left<=right)
  {
    mid=(left+right)/2;
    if(aa[mid].x==xx&&aa[mid].y==yy)
	   return true;
    else if(aa[mid].x>xx||(aa[mid].x==xx&&aa[mid].y<yy))
		right=mid-1;
	else
		left=mid+1;
  }
  return false;
}
int main()
{
  //int n;
  while(cin>>n&&n)
  {
    int i,j,count=0;
	for(i=0;i<n;++i)
	{cin>>aa[i].x>>aa[i].y;}
	sort(aa,aa+n,cmp);
	//for(i=0;i<n;++i)
	//	cout<<aa[i].x<<"  "<<aa[i].y<<endl;
	int ax,ay,bx,by;
    for(i=0;i<n;++i)
	{
	  for(j=i+1;j<n;++j)
	  {
	    if(aa[i].y==aa[j].y)
		{
		  int l=aa[j].x-aa[i].x;
		  ax=aa[i].x;
		  ay=aa[i].y-l;
		  bx=aa[j].x;
		  by=aa[j].y-l;
		  if(find(ax,ay)&&find(bx,by))
             count++;
		}
		if(aa[i].y<aa[j].y)
		{
          int dx=aa[j].x-aa[i].x;
		  int dy=aa[j].y-aa[i].y;
		  ax=aa[i].x+dy;
		  ay=aa[i].y-dx;
		  bx=aa[j].x+dy;
		  by=aa[j].y-dx;
		  if(find(ax,ay)&&find(bx,by))
			  count++;
		}
	  }
	}
	printf("%d\n",count);
  }
  return 0;
}