hdu 1240 Asteroids!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1029    Accepted Submission(s): 681

      本题是简单的广搜没有任何技巧而言，只要对这个地图进行广搜即可

代码：

 

  
    

   
     
   
      1 
   
     #include
   
     <
   
     stdio.h
   
     >
   
     

   
      2 
   
     #include
   
     <
   
     queue
   
     >
   
     

   
      3 
   
      
   
     using
   
      
   
     namespace
   
      std;

   
      4 
   
     typedef 
   
     struct
   
      node

   
      5 
   
     {

   
      6 
   
      
   
     int
   
      x,y,s;

   
      7 
   
      
   
     int
   
      time;

   
      8 
   
     }NODE;

   
      9 
   
      
   
     int
   
      dir[
   
     6
   
     ][
   
     3
   
     ]
   
     =
   
     {{
   
     -
   
     1
   
     ,
   
     0
   
     ,
   
     0
   
     },{
   
     1
   
     ,
   
     0
   
     ,
   
     0
   
     },{
   
     0
   
     ,
   
     -
   
     1
   
     ,
   
     0
   
     },{
   
     0
   
     ,
   
     1
   
     ,
   
     0
   
     },{
   
     0
   
     ,
   
     0
   
     ,
   
     -
   
     1
   
     },{
   
     0
   
     ,
   
     0
   
     ,
   
     1
   
     }};

   
     10 
   
      
   
     int
   
      main()

   
     11 
   
     {

   
     12 
   
      NODE cur,next;
   
     char
   
      map[
   
     15
   
     ][
   
     15
   
     ][
   
     15
   
     ];

   
     13 
   
      
   
     int
   
      i,j,k,n,sx,sy,ss,ex,ey,es,mark;

   
     14 
   
      
   
     char
   
      str1[
   
     10
   
     ],str2[
   
     5
   
     ];

   
     15 
   
      
   
     while
   
     (scanf(
   
     "
   
     %s%d
   
     "
   
     ,str1,
   
     &
   
     n)
   
     !=
   
     EOF)

   
     16 
   
      {

   
     17 
   
      mark
   
     =
   
     0
   
     ;

   
     18 
   
      
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     n;i
   
     ++
   
     )

   
     19 
   
      {

   
     20 
   
      
   
     for
   
     (j
   
     =
   
     0
   
     ;j
   
     <
   
     n;j
   
     ++
   
     )

   
     21 
   
      {

   
     22 
   
      getchar();

   
     23 
   
      
   
     for
   
     (k
   
     =
   
     0
   
     ;k
   
     <
   
     n;k
   
     ++
   
     )

   
     24 
   
      scanf(
   
     "
   
     %c
   
     "
   
     ,
   
     &
   
     map[i][j][k]);

   
     25 
   
      }

   
     26 
   
      }

   
     27 
   
      scanf(
   
     "
   
     %d%d%d
   
     "
   
     ,
   
     &
   
     sx,
   
     &
   
     sy,
   
     &
   
     ss);

   
     28 
   
      scanf(
   
     "
   
     %d%d%d
   
     "
   
     ,
   
     &
   
     ex,
   
     &
   
     ey,
   
     &
   
     es);

   
     29 
   
      scanf(
   
     "
   
     %s
   
     "
   
     ,str2);

   
     30 
   
      queue
   
     <
   
     NODE
   
     >
   
      qu;

   
     31 
   
      cur.x
   
     =
   
     sx;

   
     32 
   
      cur.y
   
     =
   
     sy;

   
     33 
   
      cur.s
   
     =
   
     ss;

   
     34 
   
      cur.time
   
     =
   
     0
   
     ;

   
     35 
   
      qu.push(cur);

   
     36 
   
      
   
     while
   
     (
   
     !
   
     qu.empty ())

   
     37 
   
      {

   
     38 
   
      cur
   
     =
   
     qu.front ();

   
     39 
   
      qu.pop();

   
     40 
   
      
   
     if
   
     (cur.x
   
     ==
   
     ex
   
     &&
   
     cur.y
   
     ==
   
     ey
   
     &&
   
     cur.s
   
     ==
   
     es)

   
     41 
   
      {

   
     42 
   
      mark
   
     =
   
     1
   
     ;

   
     43 
   
      
   
     break
   
     ;

   
     44 
   
      }

   
     45 
   
      
   
     for
   
     (i
   
     =
   
     0
   
     ;i
   
     <
   
     6
   
     ;i
   
     ++
   
     )

   
     46 
   
      {

   
     47 
   
      next.x
   
     =
   
     cur.x
   
     +
   
     dir[i][
   
     0
   
     ];

   
     48 
   
      next.y
   
     =
   
     cur.y
   
     +
   
     dir[i][
   
     1
   
     ];

   
     49 
   
      next.s
   
     =
   
     cur.s
   
     +
   
     dir[i][
   
     2
   
     ];

   
     50 
   
      
   
     if
   
     (next.x
   
     >=
   
     0
   
      
   
     &&
   
     next.x
   
     <
   
     n
   
     &&
   
     next.y
   
     <
   
     n
   
     &&
   
     next.y
   
     >=
   
     0
   
     &&
   
     next.s
   
     >=
   
     0
   
     &&
   
     next.s
   
     <
   
     n
   
     &&
   
     map[next.s][next.x][next.y]
   
     !=
   
     '
   
     X
   
     '
   
     )

   
     51 
   
      {

   
     52 
   
      map[next.s][next.x][next.y]
   
     =
   
     '
   
     X
   
     '
   
     ;

   
     53 
   
      next.time
   
     =
   
     cur.time
   
     +
   
     1
   
     ;

   
     54 
   
      qu.push(next);

   
     55 
   
      }

   
     56 
   
      }

   
     57 
   
      }

   
     58 
   
      
   
     if
   
     (mark
   
     ==
   
     0
   
     )

   
     59 
   
      printf(
   
     "
   
     NO ROUTE\n
   
     "
   
     );

   
     60 
   
      
   
     else
   
      
   
     if
   
     (mark
   
     ==
   
     1
   
     )

   
     61 
   
      printf(
   
     "
   
     %d %d\n
   
     "
   
     ,n,cur.time);

   
     62 
   
      }

   
     63 
   
      
   
     return
   
      
   
     0
   
     ;

   
     64 
   
     }

   
     65 
   
     

   
     66 
   
     

   
     67 
   
     

   
     68 
   
     

   
     69