[计算几何]点集中的点能组成多少个正方形

http://acm.pku.edu.cn/JudgeOnline/problem?id=2002

http://acm.pku.edu.cn/JudgeOnline/problem?id=3432

枚举正方形一条对角线线上的2个点，很容易求出另外2个点，然后要检查另外2个点是否坐标都是整数，如果都是，然后看这2个点是否在点集中，要快速的判断一个点是否在点集中，可以先把点排序，然后二分查找，或者对点的坐标进行hash处理（速度更快）。

注意点：

在检查一个浮点数(x)是否是整数时，不能这样：

int x1=(int)x;

if(fabs(x-x1)<eps) //则是整数

因为对浮点数强制类型转换的结果是丢掉小数位

比如int(1.99999999999999999)=1

而1.999999999999999可以看做就是整数2,因此在对强制类型转换得到整数a后，要分别对a和a+1与原浮点数进行做差判断！

Hash版本：

#include  < cstdio >
#include  < cmath >

const   int  maxn = 2002 ;
const   double  eps = 1.0e-6 ;
const   int  PRIME = 9991 ;

struct  Point {
     int  x,y;
};

struct  HASH {
     int  cnt;
     int  next;
} hash[ 50000 ];
int  hash1;

Point points[maxn];

inline  void  Rotate( double  x, double  y,Point P, double   & x1, double   & y1) {
     double  vx = P.x - x,vy = P.y - y;
    x1 = x;y1 = y;
    x1 -= vy;
    y1 += vx;
}

bool  getsquare(Point p1,Point p2,Point  & p3,Point  & p4) {
     double  midx = (p1.x + p2.x) / 2.0 ;
     double  midy = (p1.y + p2.y) / 2.0 ;
     double  p3x,p3y,p4x,p4y;
    Rotate(midx,midy,p1,p3x,p3y);
    Rotate(midx,midy,p2,p4x,p4y);
    p3.x = int (p3x);
    p3.y = int (p3y);
    p4.x = int (p4x);
    p4.y = int (p4y);
     bool  tag = false ;
     if (fabs(p3.x + 1 - p3x) < eps) {p3.x ++ ;tag = true ;}
     if (fabs(p3x - p3.x) < eps) tag = true ;
     if ( ! tag)  return   false ;
    tag = false ;
     if (fabs(p3.y + 1 - p3y) < eps) {p3.y ++ ;tag = true ;}
     if (fabs(p3y - p3.y) < eps) tag = true ;
     if ( ! tag)  return   false ;
    tag = false ;
     if (fabs(p4.x + 1 - p4x) < eps) {p4.x ++ ;tag = true ;}
     if (fabs(p4x - p4.x) < eps) tag = true ;
     if ( ! tag)  return   false ;
    tag = false ;
     if (fabs(p4.y + 1 - p4y) < eps) {p4.y ++ ;tag = true ;}
     if (fabs(p4y - p4.y) < eps) tag = true ;
     if ( ! tag)  return   false ;
     return   true ;
}

int  Hash( int  n) {
     int  i = n % PRIME;
     while (hash[i].next !=- 1 ) {
         if (hash[hash[i].next].cnt == n)  return   1 ;
         else   if (hash[hash[i].next].cnt > n)  break ;
        i = hash[i].next;
    }
    hash[hash1].cnt = n;
    hash[hash1].next = hash[i].next;
    hash[i].next = hash1;
    hash1 ++ ;
     return   0 ;
}

int  Hash2( int  n) {
     int  i = n % PRIME;
     while (hash[i].next !=- 1 ) {
         if (hash[hash[i].next].cnt == n)  return   1 ;
         else   if (hash[hash[i].next].cnt > n)  return   0 ;
        i = hash[i].next;
    }
     return   0 ;
}
    
int  main() {
     int  n,x,y;
    Point p1,p2;
    scanf( " %d " , & n);
         for ( int  i = 0 ;i < PRIME;i ++ ) hash[i].next =- 1 ;
        hash1 = PRIME;
         for ( int  i = 0 ;i < n;i ++ ) {
            scanf( " %d %d " , & points[i].x, & points[i].y);
            Hash((points[i].x + 100000 ) * 100000 + points[i].y + 100000 );
        }
         int  cnt = 0 ;
         for ( int  i = 0 ;i < n;i ++ ) {
             for ( int  j = i + 1 ;j < n;j ++ ) {
                 bool  ok = getsquare(points[i],points[j],p1,p2);
                 if ( ! ok)  continue ;
                 if (Hash2((p1.x + 100000 ) * 100000 + p1.y + 100000 ) == 0 )  continue ;
                 if (Hash2((p2.x + 100000 ) * 100000 + p2.y + 100000 ) == 0 )  continue ;
                cnt ++ ;
            }
        }
        printf( " %d\n " ,cnt / 2 );
     return   0 ;
}

           

排序+二分查找版本：

#include  < cstdio >
#include  < cmath >
#include  < algorithm >
using   namespace  std;

const   int  maxn = 2002 ;
const   double  eps = 1.0e-6 ;

struct  Point {
     int  x,y;
};

bool   operator < ( const  Point &  t1, const  Point &  t2) {
     if (t1.x != t2.x)  return  t1.x < t2.x;
     if (t1.y != t2.y)  return  t1.y < t2.y;
     return   false ;
}

Point points[maxn];

inline  void  Rotate( double  x, double  y,Point P, double   & x1, double   & y1) {
     double  vx = P.x - x,vy = P.y - y;
    x1 = x;y1 = y;
    x1 -= vy;
    y1 += vx;
}

bool  getsquare(Point p1,Point p2,Point  & p3,Point  & p4) {
     double  midx = (p1.x + p2.x) / 2.0 ;
     double  midy = (p1.y + p2.y) / 2.0 ;
     double  p3x,p3y,p4x,p4y;
    Rotate(midx,midy,p1,p3x,p3y);
    Rotate(midx,midy,p2,p4x,p4y);
    p3.x = int (p3x);
    p3.y = int (p3y);
    p4.x = int (p4x);
    p4.y = int (p4y);
     bool  tag = false ;
     if (fabs(p3.x + 1 - p3x) < eps) {p3.x ++ ;tag = true ;}
     if (fabs(p3x - p3.x) < eps) tag = true ;
     if ( ! tag)  return   false ;
    tag = false ;
     if (fabs(p3.y + 1 - p3y) < eps) {p3.y ++ ;tag = true ;}
     if (fabs(p3y - p3.y) < eps) tag = true ;
     if ( ! tag)  return   false ;
    tag = false ;
     if (fabs(p4.x + 1 - p4x) < eps) {p4.x ++ ;tag = true ;}
     if (fabs(p4x - p4.x) < eps) tag = true ;
     if ( ! tag)  return   false ;
    tag = false ;
     if (fabs(p4.y + 1 - p4y) < eps) {p4.y ++ ;tag = true ;}
     if (fabs(p4y - p4.y) < eps) tag = true ;
     if ( ! tag)  return   false ;
     return   true ;
}
    
int  main() {
     int  n,x,y;
    Point p1,p2;
     while (scanf( " %d " , & n) != EOF && n) {
         for ( int  i = 0 ;i < n;i ++ ) {
            scanf( " %d %d " , & points[i].x, & points[i].y);
        }
        sort(points,points + n);
         int  cnt = 0 ;
         for ( int  i = 0 ;i < n;i ++ ) {
             for ( int  j = i + 1 ;j < n;j ++ ) {
                 bool  ok = getsquare(points[i],points[j],p1,p2);
                 if (ok)
                     if (binary_search(points,points + n,p1)  &&  binary_search(points,points + n,p2)) 
                        cnt ++ ;
            }
        }
        printf( " %d\n " ,cnt / 2 );
    }
     return   0 ;
}