Squares - POJ 2002 哈希

最新推荐文章于 2018-02-04 21:40:21 发布

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本文介绍了一种通过枚举和哈希技术来寻找平面上所有可能形成的正方形的方法。输入一组坐标点，算法能高效地找出由这些点构成的所有正方形。

Squares

Time Limit: 3500MS		Memory Limit: 65536K
Total Submissions: 15556		Accepted: 5891

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

题意：这些点能够组成多少个正方形。

思路：枚举两个点，利用哈希判断另外两个点是否存在。

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;
        //printf("%d %d %d %d\n",x1,y1,x2,y2);
        if(find(x1,y1)&& find(x2,y2))
         ans++;
        x1=X[i]-y;
        y1=Y[i]+x;
        x2=X[j]-y;
        y2=Y[j]+x;
        y2=Y[j]+x;
        //printf("%d %d %d %d\n",x1,y1,x2,y2);
        if(find(x1,y1)&& find(x2,y2))
         ans++;
        // printf("%I64d\n\n",ans);
      }
    }
    printf("%I64d\n",ans/4);
  }
}