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本文介绍了一个基于广度优先搜索的3D迷宫逃逸算法实现，该算法能够找到从起点到出口的最短路径。通过使用队列来遍历迷宫，并标记已访问过的节点以避免重复计算，该算法可以有效地解决三维迷宫问题。

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!

分析：一：注意方向，有上下东西南北一共六个方向，千万不要遗漏。

二：在判断是否能走的时候 '.' 和 'E' 是或者的关系。

代码：

#include<iostream>
#include<cstdio>
#include<string.h>
#include<math.h>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<queue>
using namespace std;
const int INF = 100000000;
int mx[6] = { 0,0,1,0,-1,0 };
int my[6] = { 0,0,0,1,0,-1 };
int mz[6] = { -1,1,0,0,0,0 };
struct Point {
	int x;
	int y;
	int z;
};
int L, R, C;
char maze[30][30][30];
int ans[30][30][30];
int main()
{
	while (cin >> L >> R >> C) {
		memset(maze, 0, sizeof(maze));
		memset(ans, 0, sizeof(ans));
		int l1, l2, r1, r2, c1, c2;
		if (L == 0) {
			break;
		}
		for (int i = 0; i<L; i++) {
			for (int j = 0; j<R; j++) {
				for (int k = 0; k<C; k++) {
					cin >> maze[i][j][k];
					ans[i][j][k] = INF;
					if (maze[i][j][k] == 'S') {
						l1 = i;
						r1 = j;
						c1 = k;
					}
					if (maze[i][j][k] == 'E') {
						l2 = i;
						r2 = j;
						c2 = k;
					}
				}
			}
		}
		ans[l1][r1][c1] = 0;
		queue<Point> que;
		Point temp = { c1,r1,l1 };
		que.push(temp);
		while (que.size()) {
			Point flag = que.front();
			que.pop();
			if (maze[flag.z][flag.y][flag.x] == 'E') {
				break;
			}
			for (int i = 0; i < 5; i++) {
				int dx = flag.x + mx[i];
				int dy = flag.y + my[i];
				int dz = flag.z + mz[i];
				if (dx < C && dx >= 0 && dy < R && dy >= 0 && dz < L && ans[dz][dy][dx] == INF) {
					if (maze[dz][dy][dx] == '.' || maze[dz][dy][dx] == 'E') {
						temp = { dx,dy,dz };
						que.push(temp);
						ans[dz][dy][dx] = ans[flag.z][flag.y][flag.x] + 1;
					}
				}
			}
		}

		if (ans[l2][r2][c2] != INF)
			printf("Escaped in %d minute(s).\n", ans[l2][r2][c2]);
		else
			printf("Trapped!\n");
	}
	return 0;
}