Va 10085 the most distant state

Uva_10085

This question is basically the same as that of Digital 8, except that the digital 8 problem is the minimum step required to change to a specified image. This problem is based on a specified image, find a graph and perform the following operations to get the most steps for the specified graph.

SlaveCodeFrom the perspective of octa digital While The loop exits when a specified image is encountered. While The loop can only be Front = rear In this way, all cases are searched.

For more information about the eight digital problems, refer toP132.

 # Include  <  Stdio. h  >  
# Include  <  String  . H  > 
  # Define  Hash 100003  
  Int  St [  1000000  ] [  9  ], Fa [  1000000  ], Move [  1000000  ];
  Int  Head [  1000003 ], Next [  1000000  ];
  Int  DX []  =  {  -  1  ,  1  ,  0  ,  0  }, Dy []  =  { 0  ,  0  ,  -  1  ,  1  };
  Int  Hash (  Int     *  P)
{
  Int  I, V;
V  = 0  ;
  For  (I  =  0  ; I  <  9  ; I  ++  )
V  =  V  *  10  +  P [I];
 Return  V  %  Hash;
}
  Int  Insert (  Int  K)
{
  Int  I, h;
H  =  Hash (ST [k]);
  For  (I  =  Head [H]; I  ! =- 1  ; I  =  Next [I])
  If  (Memcmp (ST [I], St [K],  Sizeof  (St [k])  =  0  )
  Break  ;
  If  (I  =-  1 )
{
Next [k]  =  Head [H];
Head [H]  =  K;
  Return     1  ;
}
  Else  
  Return     0  ;
}
  Void  Printpath ( Int  K)
{
  If  (Fa [k]  =-  1  )
  Return  ;
Printpath (Fa [k]);
  If  (Move [k]  =  0  )
Printf (  "  U  " );
  Else     If  (Move [k]  =  1  )
Printf (  "  D  "  );
  Else     If  (Move [k]  =  2 )
Printf (  "  L  "  );
  Else  
Printf (  "  R  "  );
}
  Int  Main ()
{
  Int  I, J, K, temp, T, TT;
  Int Front, rear, X, Y, Z, newx, newy, newz;
Scanf (  "  % D  "  ,  &  T );
  For  (TT  =  0  ; TT  <  T; TT  ++  )
{
Front  = Rear  =  0  ;
  For  (I  =  0  ; I  <  3  ; I  ++  )
  For  (J  =  0 ; J  <  3  ; J  ++  )
Scanf (  "  % D  "  ,  &  St [rear] [I  *  3  +  J]);
Fa [rear]  = Move [rear]  =-  1  ;
Memset (Head,  -  1  ,  Sizeof  (Head ));
Insert (rear );
Rear  ++  ;
  While  (Front  <  Rear)
{
  For (Z  =  0  ; Z  <  9  ; Z  ++  )
  If  (St [Front] [Z]  =  0  )
  Break  ;
X  =  Z /  3  ;
Y  =  Z  %  3  ;
  For  (I  =  0  ; I  <  4  ; I  ++ )
{
Memcpy (ST [rear], St [Front],  Sizeof  (St [Front]);
Newx  =  X  +  DX [I];
Newy  =  Y  +  Dy [I];
  If  (Newx  > =  0  && Newx  <  3  &&  Newy  > =  0  &&  Newy  <  3  )
{
Newz  =  Newx  *  3  + Newy;
Temp  =  St [rear] [Z];
St [rear] [Z]  =  St [rear] [newz];
St [rear] [newz]  =  Temp;
  If  (Insert (rear ))
{
Fa [rear]  =  Front;
Move [rear]  =  I;
Rear ++  ;
}
}
}
Front  ++  ;
}
Printf (  "  Puzzle # % d \ n  "  , TT  +  1  );
  For  (I  =  0  ; I <  9  ; I  ++  )
{
  If  (I  %  3  ! =  0  )
Printf (  "     "  );
Printf (  " % D  "  , St [rear  -  1  ] [I]);
  If  (I  +  1  )  %  3  =  0  )
Printf (  "  \ N "  );
}
Printpath (Rear  -  1  );
Printf (  "  \ N  "  );
}
  Return     0  ;
}