Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28416		Accepted: 11109

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?

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.

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!

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

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

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

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

0 0 0

Escaped in 11 minute(s).
Trapped!

迷宫问题简单变形，一个3D迷宫，给你起点终点，找出最短路径。

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn =  35;
char maze[maxn][maxn][maxn];
int dis[maxn][maxn][maxn];
int L, R, C;

struct Node
{
    int z, x, y;
    Node(int zz, int xx, int yy): z(zz), x(xx), y(yy) {}
};

int bfs(int sz, int sx, int sy, int gz, int gx, int gy)
{
    queue<Node>que;
    int dx[5] = { -1, 0, 1,0}, dy[5] = {0, -1, 0, 1}, dz[3] = {0, -1, 1};
    memset(dis, INF, sizeof(dis));

    dis[sz][sx][sy] = 0;
    que.push(Node(sz, sx, sy));

    while (!que.empty())
    {
        Node pos = que.front();
        que.pop();

        if (pos.z == gz && pos.x == gx && pos.y == gy)
        {
            break;
        }

        for (int i = 0; i < 3; i++)
        {
            if (i)
            {
                int nz = pos.z + dz[i], nx = pos.x, ny = pos.y;
                if (nz >= 0 && nz < L && nx >= 0 && nx < R && ny >= 0 && ny < C && maze[nz][nx][ny] != '#' && dis[nz][nx][ny] == INF)
                {
                    que.push(Node(nz, nx, ny));
                    dis[nz][nx][ny] = dis[pos.z][pos.x][pos.y] + 1;
                }
            }
            else
            {
                for (int j = 0; j < 5; j++)
                {
                    int nz = pos.z + dz[i], nx = pos.x + dx[j], ny = pos.y + dy[j];
                    if (nz >= 0 && nz < L && nx >= 0 && nx < R && ny >= 0 && ny < C && maze[nz][nx][ny] != '#' && dis[nz][nx][ny] == INF)
                    {
                        que.push(Node(nz, nx, ny));
                        dis[nz][nx][ny] = dis[pos.z][pos.x][pos.y] + 1;
                    }
                }
            }

        }
    }
    return dis[gz][gx][gy];
}

int main()
{
    //freopen("input.txt", "r", stdin);
    while (~scanf("%d%d%d", &L, &R, &C) && L && R && C)
    {
        int sz, sx, sy, gz, gx, gy;
        for (int i = 0; i < L; i++)
        {
            for (int j = 0; j < R; j++)
            {
                scanf("%s", maze[i][j]);
                for (int k = 0; k < C; k++)
                {
                    if (maze[i][j][k] == 'S')
                    {
                        sz = i, sx = j, sy = k;
                    }
                    if (maze[i][j][k] == 'E')
                    {
                        gz = i, gx = j, gy = k;
                    }
                }
            }
            getchar();
        }
        bfs(sz, sx, sy, gz, gx, gy) == INF ? printf("Trapped!\n") : printf("Escaped in %d minute(s).\n", dis[gz][gx][gy]);
    }
    return 0;
}