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; }