Dungeon Master--POJ 2251

1、题目类型：模拟、迷宫、BFS。

2、解题思路：BFS的三维Maze[][][]应用，其每步存在前后、左右、上下6个方向的选择。

3、注意事项：BFS入队列的条件判断。

4、实现方法:

  
    

   
     
   
     #include
   
     <
   
     iostream
   
     >
   
     
#include
   
     <
   
     queue
   
     >
   
     

   
     using
   
      
   
     namespace
   
      std;


   
     struct
   
      Point
{
 
   
     int
   
      x,y,z;
 
   
     int
   
      step;
};

Point start,end;

   
     int
   
      dir[
   
     6
   
     ][
   
     3
   
     ]
   
     =
   
     {{
   
     -
   
     1
   
     ,
   
     0
   
     ,
   
     0
   
     },{
   
     1
   
     ,
   
     0
   
     ,
   
     0
   
     },{
   
     0
   
     ,
   
     -
   
     1
   
     ,
   
     0
   
     },
 {
   
     0
   
     ,
   
     1
   
     ,
   
     0
   
     },{
   
     0
   
     ,
   
     0
   
     ,
   
     -
   
     1
   
     },{
   
     0
   
     ,
   
     0
   
     ,
   
     1
   
     }};

   
     int
   
      x,y,z;

   
     char
   
      Maze[
   
     50
   
     ][
   
     50
   
     ][
   
     50
   
     ];

   
     void
   
      Init()
{
 
   
     int
   
      i,j,k;
 
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     x;i
   
     ++
   
     )
 {
 
   
     for
   
     (j
   
     =
   
     0
   
     ;j
   
     <
   
     y;j
   
     ++
   
     )
 {
 
   
     for
   
     (k
   
     =
   
     0
   
     ;k
   
     <
   
     z;k
   
     ++
   
     )
 {
 cin
   
     >>
   
     Maze[i][j][k];
 
   
     if
   
     (Maze[i][j][k]
   
     ==
   
     '
   
     S
   
     '
   
     )
 {
 start.x
   
     =
   
     i;
 start.y
   
     =
   
     j;
 start.z
   
     =
   
     k;
 }
 
   
     if
   
     (Maze[i][j][k]
   
     ==
   
     '
   
     E
   
     '
   
     )
 {
 end.x
   
     =
   
     i;
 end.y
   
     =
   
     j;
 end.z
   
     =
   
     k;
 }
 }
 }
 }
}


   
     void
   
      BFS()
{
 queue
   
     <
   
     Point
   
     >
   
      Q;
 
   
     int
   
      vis[
   
     50
   
     ][
   
     50
   
     ][
   
     50
   
     ];
 memset(vis,
   
     0
   
     ,
   
     sizeof
   
     (vis));
 start.step
   
     =
   
     0
   
     ;
 Q.push(start);
 vis[start.x][start.y][start.z]
   
     =
   
     1
   
     ;
 
   
     while
   
     (
   
     !
   
     Q.empty())
 {
 Point p
   
     =
   
     Q.front();
 Q.pop();
 
   
     if
   
     (p.x
   
     ==
   
     end.x
   
     &&
   
     p.y
   
     ==
   
     end.y
   
     &&
   
     p.z
   
     ==
   
     end.z)
 {
 cout
   
     <<
   
     "
   
     Escaped in 
   
     "
   
     <<
   
     p.step
   
     <<
   
     "
   
      minute(s).
   
     "
   
     <<
   
     endl;
 
   
     return
   
      ;
 }
 
   
     for
   
     (
   
     int
   
      i
   
     =
   
     0
   
     ;i
   
     <
   
     6
   
     ;i
   
     ++
   
     )
 {
 
   
     int
   
      x1,y1,z1;
 x1
   
     =
   
     p.x
   
     +
   
     dir[i][
   
     0
   
     ];
 y1
   
     =
   
     p.y
   
     +
   
     dir[i][
   
     1
   
     ];
 z1
   
     =
   
     p.z
   
     +
   
     dir[i][
   
     2
   
     ];
 
   
     if
   
     (x1
   
     >=
   
     0
   
     &&
   
     x1
   
     <
   
     x
   
     &&
   
     y1
   
     >=
   
     0
   
     &&
   
     y1
   
     <
   
     y
   
     &&
   
     z1
   
     >=
   
     0
   
     &&
   
     z1
   
     <
   
     z
   
     &&!
   
     vis[x1][y1][z1]
   
     &&
   
     Maze[x1][y1][z1]
   
     !=
   
     '
   
     #
   
     '
   
     )
 {
 Point tmp;
 tmp.x
   
     =
   
     x1;
 tmp.y
   
     =
   
     y1;
 tmp.z
   
     =
   
     z1;
 tmp.step
   
     =
   
     p.step
   
     +
   
     1
   
     ;
 vis[x1][y1][z1]
   
     =
   
     1
   
     ;
 Q.push(tmp);
 }
 }
 }
 cout
   
     <<
   
     "
   
     Trapped!
   
     "
   
     <<
   
     endl;
 
   
     return
   
      ;
}


   
     int
   
      main()
{
 
   
     while
   
     (cin
   
     >>
   
     x
   
     >>
   
     y
   
     >>
   
     z)
 {
 
   
     if
   
     (x
   
     ==
   
     0
   
     &&
   
     y
   
     ==
   
     0
   
     &&
   
     z
   
     ==
   
     0
   
     )
 
   
     break
   
     ;
 Init();
 BFS();
 }
 
   
     return
   
      
   
     0
   
     ;
}