題目連結：http://poj.org/problem?id=2251 Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 50755		Accepted: 19053

Description

You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides. 

Is an escape possible? If yes, how long will it take? 

Input

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size). 
L is the number of levels making up the dungeon. 
R and C are the number of rows and columns making up the plan of each level. 
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form 

Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape. 
If it is not possible to escape, print the line 

Trapped!

Sample Input

3 4 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!

Source

Ulm Local 1997 比較簡單，多輸入幾層迷宮，4個方向變成6個方向就ok了.（我也不知道為什麼，實驗課偷偷做題被這道sb題卡半天。。。可能是報復吧QAQ） 需要注意的坑點是， 輸入順序，先是z軸，再是xy的二維平面 程式碼附上：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<queue>
 4 #include<string>
 5 #include<cstring>
 6 #define sca(a) scanf("%d",&a );
 7 #define sca2(a,b) scanf("%d%d",&a,&b );
 8 #define sca3(a,b,c) scanf("%d%d%d",&a,&b,&c);
 9 #define mem(a,b) memset(a,b,sizeof(a));
10 #define FOR0(i,j) for(int i=0;i<=j;++i)
11 #define FOR1(i,j) for(int i=1;i<=j;++i)
12 using namespace std;
13 struct node
14 {
15     int x,y,z,step;
16 };
17 char plat[40][40][40];
18 int vis[40][40][40];
19 int dir[6][3]={0,0,1,0,1,0,1,0,0,-1,0,0,0,-1,0,0,0,-1};
20 int m,n,o;
21 int sx,sy,sz;
22 void ini()
23 {
24     mem(plat,0);
25     mem(vis,0);
26 }
27 int main()
28 {
29     //cout<<sx<<sy<<sz;
30     while(scanf("%d%d%d",&o,&m,&n)&&m+n+o)
31     {
32         ini();
33         for(int i=1;i<=o;i++)
34         {
35             for(int j=1;j<=m;j++)
36             {
37                 scanf("%s", plat[i][j]+1);
38                 for(int k=1;k<=n;k++)
39                 {
40                     if(plat[i][j][k]=='S')
41                     {
42                         sx=j;sy=k;sz=i;
43                     }
44                 }
45             }
46         }
47         //cout<<sx<<sy<<sz;
48         queue<node>q;
49         q.push({sx,sy,sz,0});
50         vis[sz][sx][sy]=1;
51         int flag=0;
52         while(!q.empty())
53         {
54             node now=q.front();
55             q.pop();
56             if(plat[now.z][now.x][now.y]=='E')
57             {
58                 flag=1;
59                 printf("Escaped in %d minute(s).\n", now.step);
60                 break;
61             }
62             int nex,ney,nez;
63             for(int i=0;i<6;i++)
64             {
65                 nex=now.x+dir[i][0];
66                 ney=now.y+dir[i][1];
67                 nez=now.z+dir[i][2];
68                 if(nex<1||nex>m||ney<1||ney>n||nez<1||nez>o) continue;
69                 if(vis[nez][nex][ney]) continue;
70                 if(plat[nez][nex][ney]=='.'||plat[nez][nex][ney]=='E')
71                 {
72                     //printf("%d %d %d\n",nex,ney,nez );
73                     vis[nez][nex][ney]=1;
74                     q.push({nex,ney,nez,now.step+1});
75                 }
76             }
77         }
78         if(!flag)
79             printf("Trapped!\n");
80 }
81 return 0;
82 
83 }