poj 2022 Squares

算法：

1.枚举任意两个点，根据公式计算出另外两个点坐标是否存在。

2.查找算法，可以用HASH或二分查找。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<iostream>
using namespace std;


struct Point
{       
  int x,y;
  bool operator < ( const Point &A) const
  {
    if( x != A.x )
    return x < A.x;
    return y < A.y;    
  }
}tmp[1010];


struct node
{
  Point p;
  int next;
}nt[201000];

int head[201000],size;

int find(Point A)
{
   int key = abs(A.x+A.y);
   for( int e = head[key]; e != -1; e = nt[e].next)
   {
      if( nt[e].p.x == A.x && nt[e].p.y == A.y )
          return 1;     
   }
   return 0;   
}

int main( )
{
  int N;
  while( scanf("%d", &N), N)
  {
     memset(head, -1,sizeof(head));
     size = 0;
     for( int i = 1; i <= N; i++)
     {
        scanf("%d%d",&tmp[i].x, &tmp[i].y);
        int key = abs(tmp[i].x+tmp[i].y);
        nt[size].p = tmp[i];
        nt[size].next = head[key];
        head[key] = size++;
     }
     sort( tmp + 1, tmp + N + 1);
     int num = 0;
     for( int i = 1; i <= N; i++)
     {
        for( int j = i + 1; j <= N; j++)
        {
            int a1 = tmp[i].x;
            int a2 = tmp[i].y;
            int b1 = tmp[j].x;
            int b2 = tmp[j].y;
            Point A,B;
            A.x = a1 + a2 - b2;
            A.y = a2 + b1 - a1;
            B.x = b1 + a2 - b2;
            B.y = b2 + b1 - a1;
            if( find(A) && find(B) )
                num++;
        }
         
     }
     printf("%d\n",num / 2);
  }    
    
}