题目分析

很明显的bfs，只不过三维

经验总结

没什么难点，只有一点需要注意：6个方向直接写出，开始时没搞清直接用了个for循环12个方向（肯定包括斜线了），结果样例是9步【看见部署还少了其实就该想到方向问题】
依旧强调一下语法：
g=[[[] for i in range(r)] for j in range(l)] #空行可添加
st=[[[False for i in range(c)] for j in range(r)] for k in range(l)] #空值可赋值

Python3 代码

from collections import deque

while 1:
    l,r,c=map(int,input().split())
    if l==0 and r==0 and c==0:
        break
    g=[[[] for i in range(r)] for j in range(l)]  #声明
    st=[[[False for i in range(c)] for j in range(r)] for k in range(l)]
    for i in range(l):
        for j in range(r):
            g[i][j]=list(input())
        input()
    #print(g)
    q=deque()
    dis={}
    for i in range(l):
        for j in range(r):
            for k in range(c):
                if g[i][j][k]=='S':
                    st[i][j][k]=True
                    q.append((i,j,k))
                    dis[(i,j,k)]=0
                if g[i][j][k]=='#':
                    st[i][j][k]=True
                if g[i][j][k]=='E':
                    end=(i,j,k)
    dx = [0, 0, 0, 0, 1, -1]
    dy = [0, 0, 1, -1, 0, 0]
    dz = [1, -1, 0, 0, 0, 0]
    flag=0
    while q:
        t=q.popleft()

        for i in range(6):
            nz=t[0]+dz[i]
            nx=t[1]+dx[i]
            ny=t[2]+dy[i]
            if 0<=nz<=l-1 and 0<=nx<=r-1 and 0<=ny<=c-1 and st[nz][nx][ny]==False:
                if (nz,nx,ny)==end:
                    print(f'Escaped in {dis[t]+1} minute(s).')
                    q.clear()
                    flag=1
                    break
                st[nz][nx][ny]=True
                dis[(nz,nx,ny)]=dis[t]+1
                q.append((nz,nx,ny))
                #print(nz,nx,ny)
    if flag==0:
        print("Trapped!")