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Dungeon Master
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 13923 Accepted: 5424

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!这道题最坑人的地方在于poj上将他分类为dfs，结果一写就超时，其实写最短路一般肯定是bfs，哎，经验还是太浅了。这道题还有一个值得注意的地方就是这是个3维的迷宫，只需要加上上下两个方向就行了，其他的就是简单的bfs。这道题wa了半天，居然是我忘记把文件输入输出那句话给删了，尼玛！下面是代码：

#include    #include    #define pan(a,b,c) (a<=b&&b<=c)   using namespace std;    int dir[6][3]={{0,0,-1},{0,0,1},{-1,0,0},{1,0,0},{0,-1,0},{0,1,0}};  char map[35][35][35];  int vis[35][35][35];  int t,m,n,flag;  struct node{       int x,y,z;       int num;  }que[30010];    void bfs(int sx,int sy,int sz){        int head=0;        int tail=1;        que[0].x=sx;        que[0].y=sy;        que[0].z=sz;        que[0].num=0;        vis[sx][sy][sz]=1;         flag=0;        int xx,yy,zz;        while(head>map[i][j][k];                          if(map[i][j][k]=='S'){//找到起点                                 sx=i;                                sy=j;                                sz=k;                          }                     }                     memset(vis,0,sizeof(vis));                     bfs(sx,sy,sz);       }       return 0;  }  #include 
#include 
#include 
#define pan(a,b,c) (a<=b&&b<=c)
using namespace std;

int dir[6][3]={{0,0,-1},{0,0,1},{-1,0,0},{1,0,0},{0,-1,0},{0,1,0}};
char map[35][35][35];
int vis[35][35][35];
int t,m,n,flag;
struct node{
     int x,y,z;
     int num;
}que[30010];

void bfs(int sx,int sy,int sz){
      int head=0;
      int tail=1;
      que[0].x=sx;
      que[0].y=sy;
      que[0].z=sz;
      que[0].num=0;
      vis[sx][sy][sz]=1;
       flag=0;
      int xx,yy,zz;
      while(head>map[i][j][k];
                        if(map[i][j][k]=='S'){//找到起点
                              sx=i;
                              sy=j;
                              sz=k;
                        }
                   }
                   memset(vis,0,sizeof(vis));
                   bfs(sx,sy,sz);
     }
     return 0;
}