Squares--POJ 2002

1、题目类型：枚举，哈希表。

2、解题思路：（1）建立输入点集Arr[]，并插入Hash表；（2）枚举所有线段利用Hash表寻找是否存在另外两点使其构成正方形；

3、注意事项：已知两点(a1,a2)和(b1,b2),有点(a1+(a2-b2), a2-(a1-b1))和点(b1+(a2-b2), b2-(a1-b1))可以构成一个正方形。

4、实现方法:

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     
#include
   
     <
   
     cmath
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;

   
     double
   
      
   
     const
   
      e
   
     =
   
     0.00000001
   
     ; 


   
     struct
   
      Hash{
 
   
     int
   
      num;
 Hash 
   
     *
   
     next;
 Hash()
 {
 num
   
     =-
   
     1
   
     ;
 next
   
     =
   
     NULL;
 }
 Hash(
   
     int
   
      num)
 {
 
   
     this
   
     ->
   
     num
   
     =
   
     num;
 next
   
     =
   
     NULL;
 }
};


   
     struct
   
      TPoint{
 
   
     int
   
      x,y;
};

TPoint Arr[
   
     1010
   
     ];
Hash table[
   
     40001
   
     ];


   
     //
   
     寻找点是否存在
   
     

   
     bool
   
      FindPoint(Hash table[],
   
     double
   
      x,
   
     double
   
      y)
{
 
   
     int
   
      tmp
   
     =
   
     fabs(x
   
     +
   
     y);
 
   
     if
   
     (table[tmp].num
   
     !=-
   
     1
   
     )
 {
 Hash 
   
     *
   
     p
   
     =&
   
     table[tmp];
 
   
     while
   
     (p
   
     !=
   
     NULL)
 {
 
   
     if
   
     (fabs(Arr[p
   
     ->
   
     num].x
   
     -
   
     x)
   
     <
   
     e 
   
     &&
   
      fabs(Arr[p
   
     ->
   
     num].y
   
     -
   
     y)
   
     <
   
     e)
 
   
     return
   
      
   
     true
   
     ;
 p
   
     =
   
     p
   
     ->
   
     next;
 }
 
   
     return
   
      
   
     false
   
     ;
 }
 
   
     return
   
      
   
     false
   
     ;
}


   
     int
   
      main()
{
 
   
     int
   
      n,i,j,tmp,cnt;
 
   
     while
   
     (cin
   
     >>
   
     n 
   
     &&
   
      n)
 {
 
   
     if
   
     (n
   
     ==
   
     0
   
     )
 
   
     break
   
     ;
 memset(table,NULL,
   
     sizeof
   
     (table));
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;
   
     ++
   
     i)
 {
 cin
   
     >>
   
     Arr[i].x
   
     >>
   
     Arr[i].y;
 tmp
   
     =
   
     abs(Arr[i].x
   
     +
   
     Arr[i].y); 
 
   
     if
   
     (table[tmp].num
   
     ==-
   
     1
   
     )
 table[tmp].num
   
     =
   
     i;
 
   
     else
   
      
 {
 Hash 
   
     *
   
     p
   
     =
   
     table
   
     +
   
     tmp;
 
   
     while
   
     (p
   
     ->
   
     next
   
     !=
   
     NULL)
 p
   
     =
   
     p
   
     ->
   
     next;
 p
   
     ->
   
     next
   
     =
   
     new
   
      Hash(i);
 }
 }
 cnt
   
     =
   
     0
   
     ;
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )
 
   
     for
   
     (j
   
     =
   
     i
   
     +
   
     1
   
     ;j
   
     <
   
     n;j
   
     ++
   
     )
 {
 
   
     if
   
     (FindPoint(table,(
   
     double
   
     )(Arr[i].x
   
     +
   
     Arr[j].x
   
     +
   
     Arr[j].y
   
     -
   
     Arr[i].y)
   
     /
   
     2
   
     ,
 (
   
     double
   
     )(Arr[i].x
   
     -
   
     Arr[j].x
   
     +
   
     Arr[i].y
   
     +
   
     Arr[j].y)
   
     /
   
     2
   
     )
 
   
     &&
   
      FindPoint(table,(
   
     double
   
     )(Arr[i].x
   
     +
   
     Arr[j].x
   
     -
   
     Arr[j].y
   
     +
   
     Arr[i].y)
   
     /
   
     2
   
     ,
 (
   
     double
   
     )(
   
     -
   
     Arr[i].x
   
     +
   
     Arr[j].x
   
     +
   
     Arr[i].y
   
     +
   
     Arr[j].y)
   
     /
   
     2
   
     ))
 
   
     ++
   
     cnt;
 }
 cout
   
     <<
   
     cnt
   
     /
   
     2
   
     <<
   
     endl;
 } 
 
   
     return
   
      
   
     1
   
     ;
}