POJ2251：Dungeon Master(BFS三维)

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!
 

#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
char c[40][40][40];
int vis[40][40][40];//是否访问过  
int mov[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};
int t,n,m;//t层,n行,m列 
struct node{
	int i,j,k;//i第几层，j第几行，k第几列 
	int feet;//访问到第几步 
}nd[30000];
node st,ed;
int bfs(node aa){
	aa.feet=0;
	int r=0,p=0;
	nd[p]=aa;
	vis[aa.i][aa.j][aa.k]=1;
	while(p<=r){
		node ss=nd[p];
		for(int x=0;x<6;x++){
			int ii=ss.i+mov[x][0];
			int jj=ss.j+mov[x][1];
			int kk=ss.k+mov[x][2];
			if(ii>=1&&ii<=t&&jj>=1&&jj<=n&&kk>=1&&kk<=m){
				if(!vis[ii][jj][kk]&&c[ii][jj][kk]!='#'){
					r++;
					nd[r].feet=ss.feet+1;
					nd[r].i=ii;
					nd[r].j=jj;
					nd[r].k=kk;
					vis[ii][jj][kk]=1;
					if(ii==ed.i&&jj==ed.j&&kk==ed.k){
						return nd[r].feet;
					}
				}
			}
		}
		p++;
	}
	return -1;
	
} 
int main(){
	while(~scanf("%d%d%d",&t,&n,&m)){
		if(t==0&&n==0&&m==0) break;
		memset(vis,0,sizeof(vis));
		for(int i=1;i<=t;i++){
		for(int j=1;j<=n;j++){
			for(int k=1;k<=m;k++){
				cin>>c[i][j][k];
				if(c[i][j][k]=='S'){
					st.i=i;
					st.j=j;
					st.k=k;
				}else if(c[i][j][k]=='E'){
					ed.i=i;
					ed.j=j;
					ed.k=k;
				}
			}
		}
	}
	int val=bfs(st);
	if(val==-1){
		cout<<"Trapped!"<<endl;
	}else{
		cout<<"Escaped in "<<val<<" minute(s)."<<endl;
	}
	}
	
	
	return 0;	
}