for(i=0;i<n;i++)    //输入地图
    for(j=0;j<n;j++)
        for(k=0;k<n;k++)
            cin>>a[j][k][i];

 1 #include <iostream>
 2 #include <string.h>
 3 #include <queue> 
 4 using namespace std;
 5 char a[11][11][11];    //记录地图
 6 bool isv[11][11][11];    //记录访问过没有 
 7 int dx[6] = {0,1,0,-1,0,0};
 8 int dy[6] = {1,0,-1,0,0,0}; 
 9 int dz[6] = {0,0,0,0,-1,1};
10 int n;
11 int sx,sy,sz,ex,ey,ez;
12 struct NODE{
13     int x;
14     int y;
15     int z;
16     int step;
17 };
18 bool judge(int x,int y,int z)
19 {
20     if( x<0 || y<0 || z<0 || x>=n || y>=n || z>=n )    //出界 
21         return 1;
22     if( isv[x][y][z] )    //走过 
23         return 1;
24     if( a[x][y][z]==‘X‘ )    //遇到墙 
25         return 1;
26     return 0;
27 }
28 int bfs()    //返回到达终点的时间 
29 {
30     memset(isv,0,sizeof(isv));
31     queue <NODE> q;
32     NODE cur,next;
33     cur.x = sx;
34     cur.y = sy;
35     cur.z = sz;
36     cur.step = 0;
37     isv[sx][sy][sz] = true;
38     q.push(cur);    //第一个节点入队 
39     while(!q.empty()){
40         cur = q.front();
41         q.pop();    //队首出队
42         if( cur.x==ex && cur.y==ey && cur.z==ez){
43             return cur.step;
44         }
45         for(int i=0;i<6;i++){
46             int nx = cur.x + dx[i];
47             int ny = cur.y + dy[i];
48             int nz = cur.z + dz[i];
49             if(judge(nx,ny,nz))    //判定    
50                 continue;
51             //可以走
52             next.x = nx;
53             next.y = ny;
54             next.z = nz;
55             isv[nx][ny][nz] = true;    //记录访问过 
56             next.step = cur.step + 1; 
57             q.push(next);
58         } 
59     }
60     return -1;
61 }
62 int main()
63 {
64     char str[20];
65     int i,j,k;
66     while(cin>>str>>n){
67         for(i=0;i<n;i++)    //输入地图
68             for(j=0;j<n;j++)
69                 for(k=0;k<n;k++)
70                     cin>>a[j][k][i];
71         cin>>sx>>sy>>sz;
72         cin>>ex>>ey>>ez;
73         cin>>str;
74         int step = bfs();
75         if(step!=-1)    //到达终点
76             cout<<n<<‘ ‘<<step<<endl;
77         else
78             cout<<"NO ROUTE"<<endl;
79     }
80     return 0;
81 }