Dungeon Master

最新推荐文章于 2024-09-15 11:05:32 发布

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最新推荐文章于 2024-09-15 11:05:32 发布

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原文链接：http://www.cnblogs.com/wangyumin/p/5323443.html

本文介绍了一种解决三维迷宫逃脱问题的广搜（BFS）算法，详细阐述了如何通过六方向搜索来找到从起点到终点的最短路径，并提供了具体的代码实现和输入输出示例。

算法：广搜（BFS）；

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!

三维的向六个方向搜：a[6][3]={-1,0,0,1,0,0,0,-1,0,0,1,0,0,0,-1,0,0,1};

代码：

#include <iostream>
#include <iomanip>
#include <string>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;
int n,m,k,q1,p1,s1,s2,p2,q2,t,b[35][35][35];
char map[35][35][35];
int a[6][3]={-1,0,0,1,0,0,0,-1,0,0,1,0,0,0,-1,0,0,1};
struct cm
{
	int x,y,z,step;
};
void BFS(int dx,int dy,int dz)
{   memset(b,0,sizeof(b));
	queue<cm>que ;
	cm cur,loer;
	cur.x=dx;
	cur.y=dy;
	cur.z=dz;
	cur.step=0;
	b[dx][dy][dz]=1;
	que.push(cur);
	while(que.size())
	{
		loer=que.front();
		que.pop();
		if(loer.x==s2&&loer.y==p2&&loer.z==q2)
		{
			t=1;
			cout<<"Escaped in "<<loer.step<<" minute(s)."<<endl;
			break;
		}
		for(int i=0;i<6;i++)
		{
			int nx=loer.x+a[i][0];
			int ny=loer.y+a[i][1];
			int nz=loer.z+a[i][2];
			if(nx>=0&&nx<n&&ny>=0&&ny<m&&nz>=0&&nz<k&&map[nx][ny][nz]=='.'&&!b[nx][ny][nz])
			{
				cur.x=nx;
				cur.y=ny;
				cur.z=nz;
				cur.step=loer.step+1;
				b[nx][ny][nz]=1;
				que.push(cur);
			}
		}
	}
}
int main()
{   int i,j,s;
	while(cin>>n>>m>>k&&n&&m&&k)
	{
		for(i=0;i<n;i++)
		{
			for(j=0;j<m;j++)
			{cin>>map[i][j];
				for(t=0;t<k;t++)
				{
					if(map[i][j][t]=='S')
					{
						s1=i;p1=j;q1=t;
						map[i][j][t]='.';
					}
					else if(map[i][j][t]=='E')
					{
						s2=i;p2=j;q2=t;
						map[i][j][t]='.'; 
					}
					
				}
			}
		}
		t=0;
		BFS(s1,p1,q1);
		if(t==0) cout<<"Trapped!"<<endl;
	}
}

转载于:https://www.cnblogs.com/wangyumin/p/5323443.html