http://poj.org/problem?id=2251
题意：输出从S->E的最少时间，即最少步数。
BFS搜索六个方向。
1 #include <stdio.h> 2 #include <string.h> 3 #include <queue> 4 using namespace std; 5 int e_x,e_y,e_z; 6 int n,row,col,vis[32][32][32]; 7 char map[32][32][32]; 8 int dir[6][3] = {{0,0,1},{0,0,-1},{0,-1,0},{0,1,0},{1,0,0},{-1,0,0}}; 9 struct node 10 { 11 int x; 12 int y; 13 int z; 14 int step; 15 } t,g; 16 int bfs(int x,int y,int z) 17 { 18 queue<node>q; 19 memset(vis,0,sizeof(vis)); 20 t.step = 0; 21 t.x = x; 22 t.y = y; 23 t.z = z; 24 q.push(t); 25 vis[x][y][z] = 1; 26 while(!q.empty()) 27 { 28 t = q.front(); 29 q.pop(); 30 if (t.x==e_x && t.y==e_y && t.z==e_z) 31 { 32 return t.step; 33 } 34 for (int i = 0; i < 6; i ++) 35 { 36 x = t.x +dir[i][0]; 37 y = t.y+dir[i][1]; 38 z = t.z +dir[i][2]; 39 if (map[x][y][z]!='#'&&!vis[x][y][z]) 40 { 41 g.x = x; 42 g.y = y; 43 g.z = z; 44 g.step = t.step+1; 45 q.push(g); 46 vis[x][y][z] = 1; 47 } 48 } 49 50 } 51 return 0; 52 } 53 int main() 54 { 55 56 char s[1010]; 57 while(~scanf("%d%d%d",&n,&row,&col)) 58 { 59 int i,j,k; 60 int s_x,s_y,s_z; 61 if (n==0&&row==0&&col==0) 62 break; 63 memset(map,'#',sizeof(map)); 64 for (i = 0; i < n; i ++) 65 { 66 for (j = 0; j < row; j ++) 67 { 68 scanf("%s",s); 69 70 for (k = 0 ; k < col; k ++) 71 { 72 73 if (s[k]=='S') 74 { 75 s_x = j+1; 76 s_y = k+1; 77 s_z = i+1; 78 } 79 if (s[k]=='E') 80 { 81 e_x = j+1; 82 e_y = k+1; 83 e_z = i+1; 84 } 85 map[j+1][k+1][i+1] = s[k]; 86 } 87 } 88 getchar(); 89 } 90 int ans = bfs(s_x,s_y,s_z); 91 if (ans) 92 printf("Escaped in %d minute(s). ",ans); 93 else 94 printf("Trapped! "); 95 96 } 97 return 0; 98 }