POJ2251

题目：

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!

思路：这题就是把二维的bfs扩展到3维，但是写的时候手贱，换了种判断方式，就一直超时。后面才发现这样会增加不必要的步骤。//a[cur.z][cur.x][cur.y]='#';  这样操作的话在栈中的点，也有可能到达当前的next点，
                    //若用d[next.z][next.x][next.y]==0判断，只要到达next，就不会再判断

#include <iostream>
#include<queue>
#include<algorithm>
#include<string.h>
#include<stdio.h>
using namespace std;
#define maxn 31
char a[maxn][maxn][maxn];
int d[maxn][maxn][maxn];
int dir[6][3]={{-1,0,0},{1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};//2向右3向左4向上
int sum=0;
int L,R,C;
struct node{
int x,y,z;
};
node start,ed;
int  bfs()
{
    queue<node> Q;
    Q.push(start);
    node cur;
     node next;
    while(Q.size())
    {
        cur=Q.front();
        Q.pop();
            for(int i=0;i<6;i++)
            {
                next.x=cur.x+dir[i][0];
                next.y=cur.y+dir[i][1];
                next.z=cur.z+dir[i][2];
                if(next.x>=0&&next.x<R&&next.y>=0&&next.y<C&&next.z>=0&&next.z<L&&a[next.z][next.x][next.y]!='#'&&d[next.z][next.x][next.y]==0)
                {
                    //a[cur.z][cur.x][cur.y]='#';  这样操作的话在栈中的点，也有可能到达当前的next点，
                    //若用d[next.z][next.x][next.y]==0判断，只要到达next，就不会再判断
                    d[next.z][next.x][next.y]=d[cur.z][cur.x][cur.y]+1;
                     if(next.x==ed.x&&next.y==ed.y&&next.z==ed.z)
                          return d[ed.z][ed.x][ed.y];
                     Q.push(next);
                }
            }
    }
      return -1;
}
int main()
{
   while(cin>>L>>R>>C&&(L+R+C)!=0)
   {
       sum=0;
       memset(d,0,sizeof(d));
      for(int i=0;i<L;i++)
          for(int j=0;j<R;j++)
            for(int k=0;k<C;k++)
          {

           cin>>a[i][j][k];
           if(a[i][j][k]=='S')
           {
               start.z=i;start.x=j;start.y=k;
           }
           else if(a[i][j][k]=='E')
           {
               ed.z=i;ed.x=j;ed.y=k;
           }
       }
      // cout<<start.x<< " "<<start.y<<" "<<start.z<<endl;
      // cout<<ed.x<<" "<<ed.y<<" "<<ed.z<<endl;
      if( bfs()==-1)
       printf("Trapped!\n");
      else
        printf("Escaped in %d minute(s).\n",d[ed.z][ed.x][ed.y]);

   }
    return 0;
}