POJ2251 三维迷宫（BFS

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POJ  2251 

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!

 

题目是没啥好讲的欧，就是模板从二维变三维。主要就是BFS思想：用一个队列将最优路径的每一分部存储下来。然后横向分层遍历，以记录最优路径

表达能力有限，告辞...

上code

AC:

#include 
#include 
#include 
#include 
using namespace std;
const int maxn=1000006;
struct node
{
  int x,y,z;
};
char a[34][35][36];
int n,m,k;
int sx,sy,sz;
int ex,ey,ez;
int d[34][35][36];
int dx[6]={-1,1,0,0,0,0},dy[6]={0,0,-1,1,0,0},dz[6]={0,0,0,0,1,-1};
int bfs()
{
  queueque;
  node king,_,na;
  king.x=sx;
  king.y=sy;
  king.z=sz;
  for(int i=0;i>n>>m>>k){
    if(n==0&&m==0&&k==0)break;
    for(int i=0;i>a[i][j][wyh];
  if(a[i][j][wyh]=='S')
  {
    sx=i;
    sy=j;
    sz=wyh;
  }
  else if(a[i][j][wyh]=='E')
  {
    ex=i;
    ey=j;
    ez=wyh;
  }
}
}
}
      int ans;
      ans=bfs();
        if(ans!=1000006)
        printf("Escaped in %d minute(s).\n",ans);
        else if(ans==1000006)
         printf("Trapped!\n");

}
return 0;
}