Squares - POJ 2002 哈希

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 15595		Accepted: 5901

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

题意：找到所有正方形的个数。

思路：哈希后枚举每两个点，然后看这两个点可以组成的正方形的另外两个点是不是都在。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
struct node
{ int x,y;
};
vector<node> hash[23333];
int X[1010],Y[1010],times=0;
int Hash(int x,int y)
{ int v=(x*237+y*237*237);
  if(v<0)
   v=0-v+23333;
  v%=23330;
  return v;
}
bool find(int x,int y)
{ int v=Hash(x,y);
  for(int i=0;i<hash[v].size();i++)
  { if(x==hash[v][i].x && y==hash[v][i].y)
     return true;
  }
  return false;
}
int main()
{ int n,i,j,k,v,x1,x2,y1,y2,x,y;
  long long ans;
  while(~scanf("%d",&n) && n>0)
  { times++;
    ans=0;
    memset(hash,0,sizeof(hash));
    for(i=1;i<=n;i++)
    { scanf("%d%d",&X[i],&Y[i]);
      v=Hash(X[i],Y[i]);
      node nod1;
      nod1.x=X[i];
      nod1.y=Y[i];
      hash[v].push_back(nod1);
    }
    for(i=1;i<n;i++)
    { for(j=i+1;j<=n;j++)
      { x=X[i]-X[j];
        y=Y[i]-Y[j];
        x1=X[i]+y;y1=Y[i]-x;x2=X[j]+y;y2=Y[j]-x;
        if(find(x1,y1)&& find(x2,y2))
         ans++;
        x1=X[i]-y;y1=Y[i]+x;x2=X[j]-y;y2=Y[j]+x;
        if(find(x1,y1)&& find(x2,y2))
         ans++;
      }
    }
    printf("%I64d\n",ans/4);
  }
}