C - Dungeon Master

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!

题目大意： 一个3D的地牢内有多个房间，在这个地牢内可以按照东，西，南，北，上，下方向移动，问从地牢的S房间要走到E房间最小时间花费多少。。相对与之前的广度搜
索这道题多了上下两个方向，可以定义一个三维数组map【】【】【】来存储。

代码：

#include <iostream>  
#include <stdio.h>  
#include <string.h>  
#include <queue>  
#include <algorithm>  
using namespace std;  
  
char map[35][35][35];  
int vis[35][35][35];  
int k,n,m,sx,sy,sz,ex,ey,ez;  
int to[6][3] = {{0,0,1},{0,0,-1},{0,1,0},{0,-1,0},{1,0,0},{-1,0,0}};  
  
struct node  
{  
    int x,y,z,step;  
};  
  
int check(int x,int y,int z)  
{  
    if(x<0 || y<0 || z<0 || x>=k || y>=n || z>=m)  
        return 1;  
    else if(map[x][y][z] == '#')  
        return 1;  
    else if(vis[x][y][z])  
        return 1;  
    return 0;  
}  
  
int bfs()  
{  
    int i;  
    node a,next;  
    queue<node> Q;  
    a.x = sx,a.y = sy,a.z = sz;  
    a.step = 0;  
    vis[sx][sy][sz] = 1;  
    Q.push(a);  
    while(!Q.empty())  
    {  
        a = Q.front();  
        Q.pop();  
        if(a.x == ex && a.y == ey && a.z == ez)  
            return a.step;  
        for(i = 0; i<6; i++)  
        {  
            next = a;  
            next.x = a.x+to[i][0];  
            next.y = a.y+to[i][1];  
            next.z = a.z+to[i][2];  
            if(check(next.x,next.y,next.z))  
                continue;  
            vis[next.x][next.y][next.z] = 1;  
            next.step = a.step+1;  
            Q.push(next);  
        }  
    }  
    return 0;  
}  
  
int main()  
{  
    int i,j,r;  
    while(scanf("%d%d%d",&k,&n,&m),n+m+k)  
    {  
        for(i = 0; i<k; i++)  
        {  
            for(j = 0; j<n; j++)  
            {  
                scanf("%s",map[i][j]);  
                for(r = 0; r<m; r++)  
                {  
                    if(map[i][j][r] == 'S')  
                    {  
                        sx = i,sy = j,sz = r;  
                    }  
                    else if(map[i][j][r] == 'E')  
                    {  
                        ex = i,ey = j,ez = r;  
                    }  
                }  
            }  
        }  
        memset(vis,0,sizeof(vis));  
        int ans;  
        ans = bfs();  
        if(ans)  
            printf("Escaped in %d minute(s).\n",ans);  
        else  
            printf("Trapped!\n");  
    }  
  
    return 0;  
}