poj 2251 Dungeon Master

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!



  1 #include <iostream>
  2 #include <stdio.h>
  3 #include <string.h>
  4 #include <queue>
  5 using namespace std;
  6 struct node
  7 {
  8     int x,y,z;
  9     int time;
 10 };
 11 char map[33][33][33];
 12 bool flag[33][33][33];
 13 int dir[][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};
 14 int L,R,C;
 15 int sx,sy,sz,ex,ey,ez;
 16 
 17 bool judge(node p)
 18 {
 19     if(p.x<0 || p.z<0 || p.y<0 || p.x>=L || p.y>=R || p.z>=C || flag[p.x][p.y][p.z] || map[p.x][p.y][p.z]=='#')
 20     {
 21         return false;
 22     }
 23     return true;
 24 }
 25 void bfs()
 26 {
 27     node p,temp;
 28     int i;
 29     queue<node>q;
 30     while(!q.empty())
 31     {
 32         q.pop();
 33     }
 34     flag[sx][sy][sz]=true;
 35     p.x=sx;
 36     p.y=sy;
 37     p.z=sz;
 38     p.time=0;
 39     q.push(p);
 40     while(!q.empty())
 41     {
 42         p=q.front();
 43         q.pop();
 44         for(i=0;i<6;i++)
 45         {
 46             temp=p;
 47             temp.x+=dir[i][0];
 48             temp.y+=dir[i][1];
 49             temp.z+=dir[i][2];
 50             temp.time=p.time+1;
 51             if(!judge(temp))
 52             {
 53                 continue;
 54             }
 55             if(map[temp.x][temp.y][temp.z]=='E')
 56             {
 57                 printf("Escaped in %d minute(s).\n",temp.time);
 58                 return;
 59             }
 60             flag[temp.x][temp.y][temp.z]=true;
 61             q.push(temp);
 62         }
 63     }
 64     printf("Trapped!\n");
 65 
 66 }
 67 int main()
 68 {
 69     while(~scanf("%d%d%d",&L,&R,&C))
 70     {
 71         if(L==0 && R==0 && C==0)
 72         {
 73             break;
 74         }
 75         int i,j,k;
 76         for(i=0;i<L;i++)
 77         {
 78             for(j=0;j<R;j++)
 79             {
 80                 scanf("%s",map[i][j]);
 81             }
 82         }
 83         for(i=0;i<L;i++)
 84         {
 85             for(j=0;j<R;j++)
 86             {
 87                 for(k=0;k<C;k++)
 88                 {
 89                     if(map[i][j][k]=='S')
 90                     {
 91                         sx=i;
 92                         sy=j;
 93                         sz=k;
 94                     }
 95                 }
 96             }
 97         }
 98         memset(flag,false,sizeof(flag));
 99         bfs();
100     }
101     return 0;
102 }
View Code

简单的BFS，

不过WA了一次，原因是方向数组dir[][][],的移动顺序错了，当时觉得没什么问题，就随便写的顺序，交上去没想到WA了，后来先考虑向上或者向下移动，再交，就AC了，具体什么原因我也不知道怎么解释，看到的人要是清楚，还请大神讲解

posted @ 2013-05-30 22:03  欧阳生朵  阅读(540)  评论(0)  编辑  收藏  举报