Dungeon Master

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 46 Accepted Submission(s) : 16

Problem 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<stdio.h>

#include<queue>

#include<string.h>

using namespace std;

char map[31][31][31];

int vis[31][31][31];

int dx[6]={1,0,-1,0,0,0};

int dy[6]={0,0,0,1,0,-1};

int dz[6]={0,1,0,0,-1,0};

int x,y,z,ex,ey,ez,m,n,l;

struct node

{

	int x,y,z;

	int step;

	friend bool operator< (node n1,node n2)

	{

		return n1.step>n2.step;

	}

}p,temp;

bool judge(node r)

{

	if(r.x<0||r.x>=m||r.y<0||r.y>=n||r.z<0||r.z>=l)

	return true;

	if(vis[r.x][r.y][r.z]||map[r.x][r.y][r.z]=='#')

	return true;

	return false;

}

int bfs()

{

	memset(vis,0,sizeof(vis));

	priority_queue<node>q;

	while(!q.empty()) q.pop();

	p.x=x;

	p.y=y;

	p.z=z;

	p.step=0;

	q.push(p);

	vis[x][y][z]=1;

	while(!q.empty())

	{

		p=q.top();

		q.pop();

		if(p.x==ex&&p.y==ey&&p.z==ez)

		{

			return p.step;

		}

		for(int i=0;i<6;i++)

		{

			temp=p;

			temp.x=p.x+dx[i];

			temp.y=p.y+dy[i];

			temp.z=p.z+dz[i];

			if(judge(temp))

			continue;

			vis[temp.x][temp.y][temp.z]=1;

			temp.step=p.step+1;

			q.push(temp);

		}

	}

	return 0;

}

int main()

{

	while(scanf("%d%d%d",&m,&n,&l)!=EOF)

	{

		memset(map,'\0',sizeof(map));

		getchar();

		if(m+n+l==0)

		break;

		int i,j,k;

		for(i=0;i<m;i++)

		for(j=0;j<n;j++)

		{

			scanf("%s",&map[i][j]);

			for(k=0;k<l;k++)

			{

			if(map[i][j][k]=='S')

			{

				x=i;y=j;z=k;

			}

			if(map[i][j][k]=='E')

			{

				ex=i;ey=j;ez=k;

			}

			}

		//getchar();

		}

		/*for(i=0;i<m;i++)

		for(j=0;j<n;j++)

		{

			for(k=0;k<l;k++)

			printf("%c",map[i][j][k]);

			printf("\n");

		}*/

		//printf("%d %d %d %d %d% d",x,y,z,ex,ey,ez);

		int ans=bfs();

		if(ans!=0)

		printf("Escaped in %d minute(s).\n",ans);

		else

		printf("Trapped!\n");

	}

	return 0;

}

巴特西

Dungeon Master hdoj

Dungeon Master

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 46 Accepted Submission(s) : 16

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