Acm hdu 1010 tempter of the bone

Tempter of the bone

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 23769 accepted submission (s): 6538

Problem descriptionthe doggie found a bone in an alert ent maze, which fascinated him a lot. however, when he picked it up, the maze began to shake, and the doggie cocould feel the ground sinking. he realized that the bone was a trap, and he tried desperately to get out of this maze.

The maze was a rectangle with sizes N by M. there was a door in the maze. at the beginning, the door was closed and it wocould open at the T-th second for a short period of time (less than 1 second ). therefore the doggie had to arrive at the door on exactly the T-th second. in every second, he cocould move one block to one of the upper, lower, left and right neighboring blocks. once he entered a block, The ground of this block wocould start to sink and disappear in the next second. he coshould not stay at one block for more than one second, nor coshould he move into a visited Block. can the Poor doggie keep ve? Please help him.

Inputthe input consists of multiple test cases. the first line of each test case contains three integers n, m, and T (1 <n, m <7; 0 <t <50 ), which denote the sizes of the maze and the time at which the door will open, respectively. the next n lines give the maze layout, with each line containing M characters. A character is one of the following:

'X': a block of wall, which the doggie cannot enter;
'S ': the start point of the doggie;
'D': the door; or
'.': An empty block.

The input is terminated with three 0's. This test case is not to be processed.

Outputfor each test case, print in one line "yes" if the doggie can have ve, or "no" otherwise.

Sample input4 4 5
S.x.
. X.
.. XD
....
3 4 5
S.x.
. X.
... D
0 0 0

Sample outputno
Yes

Authorzhang, Zheng

Sourcezjcpc2004

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 # Include  <  Iostream  >  
# Include  <  Queue  > 
# Include  <  String  . H  >  
# Include  <  Math. h  >  
  Using     Namespace  STD;

  Struct  Node
{
  Int X, Y, T;
};
  Int  DX []  =  {  1  ,  0  ,  -  1  ,  0  };
  Int  Dy []  =  { 0  ,  1  ,  0  ,  -  1  };
  Int  SX, Sy;
  Int  N, m, T;
  Char  Map [  10  ] [  10 ];

  Bool  CNT;
  Int  Num;
  Void  DFS (node P)
{
  If  (CNT)  Return  ;
  If  (P. x  =  SX  &&  P. Y =  Sy  &&  P.t  =  T) {CNT  =  True  ;  Return  ;}
  If  (P. t  > =  T)  Return  ;
  Int Min  =  (  Int  ) (FABS ((  Double  ) (P. x  -  SX ))  +  FABS ((  Double  ) (P. Y  -  Sy )));
  If  (Min  > T  -  P. t)  Return  ;
  If  (Min  %  2  ! =  (T  -  P. t)  %  2  )  Return  ;
 //  If (Map [SX + 1] [sy]! = '.' & Map [SX-1] [sy]! = '.' & Amp; Map [SX] [SY + 1]! = '.' & Map [SX] [SY-1]! = '.') Return;  
    If  (P. x  =  SX  &&  P. Y  =  Sy)  Return  ;
Node TMP;
  For  ( Int  I  =  0  ; I  <  4  ; I  ++  )
{
TMP. x  =  P.x  +  DX [I];
TMP. Y  =  P. Y  + Dy [I];
TMP. t  =  P.t  +  1  ;
  If  (CNT  |  TMP. x  <  1  |  TMP. x  >  N  |  TMP. Y <  1  |  TMP. Y  >  M  |  Map [TMP. x] [TMP. Y]  =  '  X  '  )  Continue  ;
  If  (TMP. x  = SX  &&  TMP. Y  =  Sy  &&  TMP. t  =  T) {CNT  =  True  ;  Return  ;}
  Char  Ch  = Map [TMP. x] [TMP. Y];
Map [TMP. x] [TMP. Y]  =  '  X  '  ;
DFS (TMP );
Map [TMP. x] [TMP. Y]  =  Ch;

}
}
  Int  Main ()
{
Node P;
  While  (Scanf (  "  % D  " ,  &  N,  &  M,  &  T)  ! =  EOF)
{
  If  (N  =  0  &&  M  =  0  && T  =  0  )  Break  ;
Memset (map,  0  ,  Sizeof  (MAP ));
Num  =  0  ;
  For  (  Int  I =  1  ; I  <=  N; I  ++  )
  For  (  Int  J  =  1  ; J  <=  M; j  ++  )
{
CIN >  Map [I] [J];

  If  (Map [I] [J]  =  '  S  '  )
{P. x  =  I; p. Y  =  J; P. t  =  0  ;}
  Else    If  (Map [I] [J]  =  '  D  '  )
{SX  =  I; SY  =  J ;}
  Else  Num  ++  ;
}
  If (Num  <  T  -  1  ) {Cout  <  "  No  "  <  Endl;  Continue  ;}
Map [p. x] [P. Y]  =  '  X '  ;
CNT  =  False  ;
DFS (P );
  If  (CNT) cout  <  "  Yes  "  <  Endl;
  Else  Cout  <  " No  "  <  Endl;
}
  Return     0  ;
}