Dungeon Master 三维搜索(bfs)

最新推荐文章于 2023-03-19 16:09:05 发布

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本文介绍了一种解决三维迷宫问题的算法实现，通过广度优先搜索（BFS）找到从起点到出口的最短路径。文章详细展示了如何利用队列数据结构进行节点遍历，并介绍了如何判断迷宫中每个单元的状态。

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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 <blockquote>Escaped in x minute(s). </blockquote> where x is replaced by the shortest time it takes to escape. If it is not possible to escape, print the line <blockquote>Trapped! </blockquote>

Sample Input

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

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

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

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

0 0 0

Sample Output

  
   Escaped in 11 minute(s).
Trapped!

#if 0
#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
int l,c,r,step[50][50][50],ei,ej,ek,si,sj,sk;
char a[50][50][50];
bool flag[50][50][50],flag1;

int f[6][3]={{1,0,0},{-1,0,0},{0,1,0},{0,-1,0},{0,0,1},{0,0,-1}};

int ok(int k,int i, int j)
{
	if(k>=0&&k<l&&i>=0&&i<r&&k>=0&& k<c) 
		return 1;
	else
		return 0;	
}

void bfs(int z,int x,int y)
{
	queue<int> q;	
	q.push(z);
	q.push(x);
	q.push(y);
	
   	while(!q.empty()) 
	{
	
		int ai,aj,ak;
		ak=q.front();
		q.pop();
		ai=q.front();
		q.pop();
		aj=q.front();
		q.pop();
		
		if(ak==ek && ai==ei && aj==ej)
		{
			cout<<step[ak][ai][aj]<<endl;
			flag1=1;
			return;
		}
		
		for(int i=0; i<6; i++) 
		{
			int	ki=ak+f[i][0];
			int ii=ai+f[i][1];
			int ji=aj+f[i][2];
			if(ki==ek && ii==ei && ji==ej)
			{
				cout<<"Escaped in "<<step[ak][ai][aj]+1<<" minute(s)."<<endl;
				flag1=1;
				return;
			}
            
			if(a[ki][ii][ji]=='.' && ok(ki,ii,ji) && flag[ki][ii][ji]==0)
			{
				flag[ki][ii][ji]=1;
				q.push(ki);
				q.push(ii);
				q.push(ji);
				step[ki][ii][ji]=step[ak][ai][aj]+1;
			}
		}
	}
	
}


int main()
{
	
	while(cin>>l>>r>>c && l+r+c)
	{		
		memset(step,0,sizeof(step));
		memset(a,0,sizeof(a));
		memset(flag,0,sizeof(flag));
		flag1=0,ek=0,ei=0,ej=0,si=sj=sk=0;
		
		for(int k=0; k<l; k++)
			for(int i=0; i<r; i++)
				for(int j=0; j<c; j++)
				{
					cin>>a[k][i][j];
					if(a[k][i][j]=='S')
					{
						sk=k;
						si=i;
						sj=j;
					}
					
					if(a[k][i][j]=='E')
					{
						ek=k;
						ei=i;
						ej=j;
					}
					
				 } 
		flag[sk][si][sj]=1;		 
		bfs(sk,si,sj);
		
		if(flag1==0)
			cout<<"Trapped!"<<endl;
		
	}
	
}
#endif