POJ

Squares

Time Limit: 3500MS Memory Limit: 65536K
Total Submissions: 20520 Accepted: 7887

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

Rocky Mountain 2004

题意：

给定n个个点，问能组成多少个矩形

n<=1000，很容易想到，枚举一条边，每条边最多是两个不同矩形的边，每次把边顺时针转90，逆时针转90去找其他两个点是否存在

判是否存在可以利用set，元素为点，每次去find即可（可能会T，程序3485ms爬过。。稳得不对。。时限3.5s=3500ms）

也可以哈希，key=(100001*x+y)%mod，再加挂链哈希

更容易想到的，如果直接暴力就开bool数组判vis[X][Y]是否存在，那么我们只需要把x，y离散化，最多只有n个值，每次计算出其他正方形的点，去离散完的横纵坐标找新的X’,Y’，然后直接vis[X’][Y’]即可时间复杂度O(N^2logN),应该是最优且不会退化的了，有空再更离散的代码

#include<cstdio>#include<algorithm>#include<cstring>#include<set>#include<cmath>using namespace std;const int maxn=1000+5;//typedef pair<int,int> pai;struct pai{  int x,y;  bool operator <(const pai&B)const   {    return (x<B.x)||(x==B.x&&y<B.y);  }}a[maxn];int ans,n;set<pai>Q;int main(){  while(~scanf("%d",&n))  {    if(!n)break;    ans=0;    Q.clear();    for(int i=1;i<=n;++i)    {      scanf("%d%d",&a[i].x,&a[i].y);      Q.insert(a[i]);    }    for(int i=1;i<=n;++i)      for(int j=i+1;j<=n;++j)      {        pai tmp,n1,n2;        tmp.x=a[i].x-a[j].x;        tmp.y=a[i].y-a[j].y;        n1.x=a[i].x-tmp.y;        n1.y=a[i].y+tmp.x;        n2.x=a[j].x-tmp.y;        n2.y=a[j].y+tmp.x;        if((Q.find(n1)!=Q.end())&&(Q.find(n2)!=Q.end()))++ans;        tmp.x=-a[i].x+a[j].x;        tmp.y=-a[i].y+a[j].y;        n1.x=a[i].x-tmp.y;        n1.y=a[i].y+tmp.x;        n2.x=a[j].x-tmp.y;        n2.y=a[j].y+tmp.x;        if((Q.find(n1)!=Q.end())&&(Q.find(n2)!=Q.end()))++ans;      }    printf("%d\n",ans/4);  }  return 0;}