1727: Dungeon Master

1727: Dungeon Master

描述

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? 

输入

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.

输出

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!

样例输入

3 4 5
S....
.###.
.##..
###.#

#####
#####
##.##
##...

#####
#####
#.###
####E

1 3 3
S##
#E#
###

0 0 0

样例输出

Escaped in 11 minute(s).
Trapped!

#include<stdio.h>
#include<string.h>

typedef struct ho{
	int x;
	int y;
	int z;
	int len;
}hole;

hole Q[30000],s,e;
int f[40][40][40];
int H,L,W;//L,R,C
int a[6][3]={{-1,0,0},{0,1,0},{1,0,0},{0,-1,0},{0,0,-1},{0,0,1}};//6个方向 

void bfs(){
	int i,j,k=1,top=1,x1,y1,z1,find=0;
	Q[k].x=s.x;//起点入队列 
	Q[k].y=s.y;
	Q[k].z=s.z;
	Q[k].len=0; 
	f[s.x][s.y][s.z]=-1;
	while(top<=k&&!find){//队列不为空 
		for(i=0;i<6;i++){
			x1=Q[top].x+a[i][0];
			y1=Q[top].y+a[i][1];
			z1=Q[top].z+a[i][2];
			if(x1==e.x&&y1==e.y&&z1==e.z){//找到出口 
				find=1;
				printf("Escaped in %d minute(s).\n",Q[top].len+1);//输出最短路径长度 
				break;
			}
			if(f[x1][y1][z1]==1){//不是出口，入队列 
				k++;
				Q[k].x=x1;
	            Q[k].y=y1;
	            Q[k].z=z1;
	            Q[k].len=Q[top].len+1;
				f[x1][y1][z1]=-1;	
			}
		}
		top++;
	}
	if(find==0){
		printf("Trapped!\n");
	}
}

int main(){
	char c;
	int i,j,k;
	while(scanf("%d%d%d",&H,&L,&W)&&(H!=0 ||L!=0 || W!=0)){
	  memset(f,0,sizeof(f));
	  getchar();
	  for(k=1;k<=H;k++){
		for(i=1;i<=L;i++){
			for(j=1;j<=W;j++){
				scanf("%c",&c);
				if(c=='S'){
					s.x=i;
					s.y=j;
					s.z=k;
				}
				else if(c=='E'){
					e.x=i;
					e.y=j;
					e.z=k;
				}
				else if(c=='.'){
					f[i][j][k]=1;
				}				
			}
			getchar();
		}
		if(k!=H){
			getchar();
		} 
	 }
	  bfs();
    }
	return 0;
}