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本文介绍了一种解决三维迷宫问题的BFS算法实现，旨在寻找从起点到终点的最短路径。若路径存在，则输出所需最短时间；若不存在，则输出无法逃脱的信息。

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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!

这道题的基本题意为让你从一个点到另一个点，若能到达则输出最低步数，若不能到达则输出Trapped!

这是一道BFS题，从所给点开始向前后左右上下六个方向搜索，直到达到要求的点或直到无路可走。

源代码如下：

#include<cstdio>
#include<cstring>
#include<queue>
#include<iostream>
using namespace std;
 
int dx[6] = {0,0,0,0,1,-1};
int dy[6] = {0,0,1,-1,0,0};
int dz[6] = {1,-1,0,0,0,0};
 
int vis[40][40][40], Map[40][40][40];
struct node
{
    int x, y, z, step;
}st, ed;
queue<node> Q;
 
bool judge(int x, int y, int z) 
{
    if(!Map[x][y][z] || vis[x][y][z])
        return false;
    return true;
}
 
int BFS()
{
    int x, y, z, t, i;
    while(!Q.empty())
    {
        node tmp = Q.front();
        Q.pop();
        x = tmp.x;
        y = tmp.y;
        z = tmp.z;
        t = tmp.step;
        for(i = 0; i < 6; i++)
        {
            int nx = x + dx[i];
            int ny = y + dy[i];
            int nz = z + dz[i];
            if(judge(nx,ny,nz))
            {
                 if(nx == ed.x && ny == ed.y && nz == ed.z)
                    return t + 1;
                 vis[nx][ny][nz] = 1;
                 node temp;
                 temp.x = nx;
                 temp.y = ny;
                 temp.z = nz;
                 temp.step = t + 1;
                 Q.push(temp);
            }
        }
    }
    return -1;
}
 
int main()
{
    int L, R, C, i, j, k;
    char ch;
    while(~scanf("%d%d%d",&L, &R, &C) && (L + R + C))
    {
        while(!Q.empty())  Q.pop();
        memset(vis,0,sizeof(vis));
        memset(Map,0,sizeof(Map));
        for(i = 1; i <= L; i++)
            for(j = 1; j <= R; j++)
                for(k = 1; k <= C; k++)
                {
                    cin >> ch;
                    if(ch == '.')
                        Map[j][k][i] = 1;
                    else if(ch == 'S')
                    {
                        st.x = j;
                        st.y = k;
                        st.z = i;
                        st.step = 0;
                        vis[j][k][i] = 1;
                        Map[j][k][i] = 1;
                        Q.push(st);
                    }
                    else if(ch == 'E')
                    {
                        ed.x = j;
                        ed.y = k;
                        ed.z = i;
                        Map[j][k][i] = 1;
                    }
                }
        int ans = BFS();
        if(ans == -1)
            printf("Trapped!\n");
        else
            printf("Escaped in %d minute(s).\n",ans);
    }
    return 0;
}