POJ 2251 Dungeon Master(三维bfs)

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20940		Accepted: 8118

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!

 

 

 

题意：你被困在一个3D的地牢中，给出地牢地图，#表示岩石，.表示道路，S表示人所在的位置，E表示出口。问是否能跑到出口，若能，输出最小时间。

 

很基础的bfs，就是加上三维地图的概念，用三维数组解决就可以了，代码如下：

 

<span style="font-size:12px;">#include<cstdio>
#include<queue>
using namespace std;
int L,R,C,ans,z,x,y,ez,ex,ey,mark[32][32][32];
char map[32][32][32];
int dir[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};

struct node
{
	int z,x,y,step;
}a,temp;

int ok(node temp)
{
	if(temp.z<0||temp.z>=L)
	   return 0;
	if(temp.x<0||temp.x>=R)
	   return 0;
	if(temp.y<0||temp.y>=C)
	   return 0;
	if(map[temp.z][temp.x][temp.y]=='#')
	   return 0;
	if(temp.step>=ans)
	   return 0;
	return 1;
}

void bfs()
{
	int i;
	a.z=z;
	a.x=x;
	a.y=y;
	a.step=0;
	queue<node>q;
	q.push(a);
	while(!q.empty())
	{
		a=q.front();
		q.pop();
		for(i=0;i<6;i++)
		{
			temp.z=a.z+dir[i][0];
			temp.x=a.x+dir[i][1];
			temp.y=a.y+dir[i][2];
			temp.step=a.step+1;
			if(ok(temp))
			{
				if(temp.z==ez&&temp.x==ex&&temp.y==ey)
				    ans=temp.step;
				map[temp.z][temp.x][temp.y]='#';
				q.push(temp);
			}
		}
	}
}

int main()
{
	int i,j,k;
	while(scanf("%d%d%d",&L,&R,&C)&&L||R||C)
	{
		for(i=0;i<L;i++)
		{
			for(j=0;j<R;j++)
			{
				scanf("%s",map[i][j]);
				for(k=0;k<C;k++)
				{
					if(map[i][j][k]=='S')
					{
						z=i;
						x=j;
						y=k;
					}
					if(map[i][j][k]=='E')
					{
						ez=i;
						ex=j;
						ey=k;
					}
				}
			}
		}
		ans=30000;
		bfs();
		if(ans==30000)
		   printf("Trapped!\n");
		else
		   printf("Escaped in %d minute(s).\n",ans);
	}
	return 0;
}</span>