hdu 1240 3维迷宫 求起点到终点的步数 (BFS)

题意，给出一个N，这是这个三空间的大小，然后给出所有面的状况O为空地,X为墙，再给出起始点的三维坐标和终点的坐标，输出到达的步数

比较坑 z是x，x是y，y是z，
Sample Input
START 1
O
0 0 0
0 0 0
END
START 3
XXX
XXX
XXX
OOO
OOO
OOO
XXX
XXX
XXX
0 0 1
2 2 1
END
START 5
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
XXXXX
XXXXX
XXXXX
XXXXX
XXXXX
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
OOOOO
0 0 0
4 4 4
END

Sample Output
1 0
3 4
NO ROUTE

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <queue>
 using namespace std;

 char map[][][];
 int v[][][];
 int n;
 int sx,sy,sz;
 int ex,ey,ez;
 int dx[] = {,-,,,,};
 int dy[] = {,,,-,,};
 int dz[] = {,,,,,-};
 int ans ;

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

 int bfs()
 {
     node now , t ;
     int fx , fy , fz ;
     queue<node> q ;
     now.x = sx ;
     now.y = sy ;
     now.z = sz ;
     now.step =  ;
     memset(v,,sizeof(v)) ;
     q.push(now) ;
     v[sz][sx][sy] =  ;
     while(!q.empty())
     {
         now = q.front() ;
         q.pop() ;
         if (now.x == ex && now.y == ey && now.z == ez)
         {
             ans  = now.step ;
             return   ;
         }
         for (int i =  ; i <  ; i++)
        {
         fx = now.x + dx[i] ;
         fy = now.y + dy[i] ;
         fz = now.z + dz[i] ;
         if (fx< || fy< || fz< || fx>=n || fy>=n || fz>=n || v[fz][fx][fy] ==  || map[fz][fx][fy] == 'X')
             continue ;

         t.x = fx ;
         t.y = fy ;
         t.z = fz ;
         t.step = now.step +  ;
         v[fz][fx][fy] =  ;
         q.push(t) ;
         }
     }

     return  ;
 }

 int main ()
 {
     //freopen("in.txt","r",stdin) ;
     char s[] ;
     while (scanf("%s %d" , s , &n) !=EOF)
     {
         int i , j , k ;
         for (i =  ; i < n ; i ++)
             for (j =  ; j < n ; j ++)
                   {
                       scanf("%s" , map[i][j]) ;
                   }
         scanf("%d %d %d" , &sx,&sy,&sz) ;
         scanf("%d %d %d" , &ex,&ey,&ez) ;
         scanf("%s" , s) ;

         ans =  ;
         if (bfs())
             printf("%d %d\n" , n , ans) ;
         else
             printf("NO ROUTE\n") ;
     }

     return  ;
 }