POJ 2251 Dungeon Master （三维BFS）

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!

这就是一个三维BFS模板题，这里需要建议处理几个小细节，就是三维BFS有六个方向，如果怕被写错，debug起来很麻烦的话，建议把二维数组分开来写，也就是写成三个一维数组一一对应，这样看起来就方便许多，另外也可以把判断函数放在外面
BFS模板题有蛮多个，像简单BFS，三维BFS，双向BFS，三向BFS，这些我都会一一补齐，毕竟会了模板才会更难的题目，下面给出链接
简单BFS
双向BFS
以及这篇文章的三维BFS，三向BFS日后更新
AC代码

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