ACM刷题之POJ————Dungeon Master

最新推荐文章于 2022-11-02 20:38:04 发布

小时的枫

最新推荐文章于 2022-11-02 20:38:04 发布

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分类专栏： ACM题目搜索文章标签： bfs ACM poj

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本文介绍了一个3D迷宫问题的解决方案，利用宽度优先搜索(BFS)算法寻找从起点到出口的最短路径。通过遍历每个单元格来确定可行路线，并最终输出到达出口所需的最短时间。

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Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30841		Accepted: 11947

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!

y一道简单的3D的BFS题目

注意题目中的Ｓ和Ｅ的位置是不确定的

用三维数组记录地图，同时Vis跟上就好

下面是ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<string>
#include<iostream>
using namespace std;
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N=2e5+7;

struct bs
{
	int x,y,l,bu;
};

int l,r,c;

bs fir;
char maps[100][100][100];
int vis[100][100][100];

void bfs(bs a)
{
	queue<bs>q;
	bs temp1,temp2;
	
	q.push(a);
	
	while(!q.empty())
	{
		temp1=q.front();
		//printf("%d %d %d\n",temp1.l,temp1.x,temp1.y);
		if(maps[temp1.l][temp1.x][temp1.y]=='E')
		{
			break;
		}
		q.pop();
		//++temp1.bu;
		
		temp2=temp1;
		++temp2.x;
		if(temp2.x<r&&vis[temp2.l][temp2.x][temp2.y]==0&&maps[temp2.l][temp2.x][temp2.y]!='#')
		{
			++temp2.bu;
			q.push(temp2);
			vis[temp2.l][temp2.x][temp2.y]=1;
		}
		
		temp2=temp1;
		++temp2.y;
		if(temp2.y<c&&vis[temp2.l][temp2.x][temp2.y]==0&&maps[temp2.l][temp2.x][temp2.y]!='#')
		{
			++temp2.bu;
			q.push(temp2);
			vis[temp2.l][temp2.x][temp2.y]=1;
		}
		
		temp2=temp1;
		++temp2.l;
		if(temp2.l<l&&vis[temp2.l][temp2.x][temp2.y]==0&&maps[temp2.l][temp2.x][temp2.y]!='#')
		{
			++temp2.bu;
			q.push(temp2);
			vis[temp2.l][temp2.x][temp2.y]=1;
		}
		
		temp2=temp1;
		--temp2.x;
		if(temp2.x>=0&&vis[temp2.l][temp2.x][temp2.y]==0&&maps[temp2.l][temp2.x][temp2.y]!='#')
		{
			++temp2.bu;
			q.push(temp2);
			vis[temp2.l][temp2.x][temp2.y]=1;
		}
		
		temp2=temp1;
		--temp2.y;
		if(temp2.y>=0&&vis[temp2.l][temp2.x][temp2.y]==0&&maps[temp2.l][temp2.x][temp2.y]!='#')
		{
			++temp2.bu;
			q.push(temp2);
			vis[temp2.l][temp2.x][temp2.y]=1;
		}
		
		temp2=temp1;
		--temp2.l;
		if(temp2.l>=0&&vis[temp2.l][temp2.x][temp2.y]==0&&maps[temp2.l][temp2.x][temp2.y]!='#')
		{
			++temp2.bu;
			q.push(temp2);
			vis[temp2.l][temp2.x][temp2.y]=1;
		}
	}
	if(temp1.bu==0)
	{
		printf("Trapped!\n");
	}else
	{
		printf("Escaped in %d minute(s).\n",temp1.bu);
	}
	
}

int main()
{
	//freopen("f:/input.txt", "r", stdin);
	int fx,fy,fl,f;
	while(scanf("%d%d%d",&l,&r,&c)!=EOF&&l&&r&&c)
	{
		CLR(vis,0);
		f=0;
		for(int i=0;i<l;i++)
		{
			getchar();
			for(int j=0;j<r;j++)
			{
				scanf("%s",maps[i][j]);
				for(int x=0;x<c&&f==0;x++)
					if(maps[i][j][x]=='S')
					{
						fir.x=j;
						fir.y=x;
						fir.l=i;
						fir.bu=0;
						f=1;
						break;
					}
			}
				
		}
		getchar();
		vis[fir.l][fir.x][fir.y]=1;
		bfs(fir);
			
		
	}
}