pku 2002 -- Squares(二分,几何)

 SquaresTime Limit: 3500MS Memory Limit: 65536K
Total Submissions: 8787 Accepted: 3051


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

 

 

 

////////////////////////////////////////////////////////////////////////// // 184K 1688MS // 公式推导 : // 如图所示,正方形的四个点分别为 (x1,y1),(x2,y2),(x3,y3),(x4,y4) // 假设一开始给出的点为 (x1,y1),(x2,y2) 且已排序 // 根据全等三角形,可以得知2个蓝色△全等，2个红色△全等 (注:此处也可推右边的正方形) // 所以可以推出 x3=x2-(y2-y1); // y3=y2+(x2-x1); // x4=x1-(y2-y1); // y4=y1+(x2-y1); // 当然全体是排好序。 因为这样子做会导致边被重用，所以需除于 2 ，具体追究可自己详细思考 ////////////////////////////////////////////////////////////////////////// #include<iostream> using namespace std; struct point { int x,y; point(int a,int b) { x=a; y=b; } point(){} }num[1010]; int n; int cmp(const void *a,const void *b) { if((*(point*)a).x==(*(point*)b).x) return (*(point*)a).y-(*(point*)b).y; return (*(point*)a).x-(*(point*)b).x; } bool next(point a,point b) //二分查找的比较函数 { if(a.x==b.x) return a.y<b.y; return a.x<b.x; } bool mysearch(point p) //二分查找 { int mid,high=n,low=0; while(low<=high) { mid=(low+high)/2; if(num[mid].x==p.x&&num[mid].y==p.y) return true; else if(next(num[mid],p)) low=mid+1; else high=mid-1; } return false; } int main() { int i,j,x3,y3,x4,y4; while(scanf("%d",&n)&&n) { for(i=0;i<n;i++) scanf("%d%d",&num[i].x,&num[i].y); qsort(num,n,sizeof(num[0]),cmp); int ans=0; for(i=0;i<n;i++) for(j=i+1;j<n;j++) { x3=num[j].x-(num[j].y-num[i].y); y3=num[j].y+(num[j].x-num[i].x); if(!mysearch( point(x3,y3) )) continue; x4=num[i].x-(num[j].y-num[i].y); y4=num[i].y+(num[j].x-num[i].x); if(!mysearch( point(x4,y4) )) continue; ans++; } printf("%d/n",ans/2); } }