poj 2251 BFS

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最新推荐文章于 2022-03-16 12:53:47 发布

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本文探讨了深度学习在图像处理领域的应用，包括图像色彩空间的理解、视频编解码算法、音频编解码算法、视频容器技术以及FFmpeg音视频编解码等关键点。此外，还涉及了图像剪辑、视频特效、动画制作和OpenCV图像处理等高级应用。

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Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21312		Accepted: 8283

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!

Source

Ulm Local 1997

给出一三维空间的地牢,要求求出由字符'S'到字符'E'的最短路径

移动方向可以是上，下，左，右，前，后，六个方向

每移动一次就耗费一分钟，要求输出最快的走出时间。

用BFS求最短路比DFS好;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>

using namespace std;

struct node
{
	int x,y,z;
}p,s,e,q;
queue<node>Q;
int n,m,h;
char a[100][100][100];  //三维
int vis[100][100][100];
int dx[]={1,-1,0,0,0,0};
int dy[]={0,0,1,-1,0,0};
int dz[]={0,0,0,0,-1,1};
int step[100][100][100];
int Judge(node q)
{
	if(q.x>=0&&q.y>=0&&q.z>=0&&q.x<n&&q.y<m&&q.z<h&&a[q.x][q.y][q.z]!='#') //存在E
		return 1;
	return 0;
}
int BFS()
{
	while(!Q.empty())   //  <span id="transmark"></span>队列必须清空
		Q.pop();
	
	Q.push(s);
	vis[s.x][s.y][s.z]=1;
	step[s.x][s.y][s.z]=0;
	while(!Q.empty())
	{
		p=Q.front();
		Q.pop();
		for(int i=0;i<6;i++)
		{
			q.x=dx[i]+p.x;
			q.y=dy[i]+p.y;
			q.z=dz[i]+p.z;
			if(!Judge(q))
				continue;
			if(vis[q.x][q.y][q.z]==0)
			{
				Q.push(q);
				step[q.x][q.y][q.z]=step[p.x][p.y][p.z]+1;
				vis[q.x][q.y][q.z]=1;
			}
			if(a[q.x][q.y][q.z]=='E')
				return step[q.x][q.y][q.z];
		}
	}
	return -1;
}
int main()
{
	while(~scanf("%d%d%d",&n,&m,&h))
	{
		int flag=1;
		if(!n&&!m&&!h)
		break;
		memset(vis,0,sizeof(vis));
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				scanf("%s",a[i][j]);
				if(flag)
				for(int k=0;k<h;k++)
				{
					if(a[i][j][k]=='S')
					{
						s.x=i;s.y=j;s.z=k;
						flag=0;
						break;
					}
				}
			}
		}
		int x=BFS();
		if(x==-1)
		{
			printf("Trapped!\n");
	        continue;
		}
		printf("Escaped in %d minute(s).\n",x);
	}
}