POJ-2251___走迷宫——解题报告 BFS

原题地址：http://poj.org/problem?id=2251

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 41530		Accepted: 15736

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!

题目大意：

给出一个三维空间迷宫，“.”代表可行走的空间，“#”代表不可行走的封闭间。从"S"开始，求到地点“E”的最短时间。

每次移动可在迷宫内上下左右前后移动。

解题思路：

BFS即可

代码思路：

1、用结构体存储长宽高和时间

2、命名2个开始与结束的结构体、一个将要执行的顺序体queue与一个正在执行的结构体x

3、三维遍历未访问过的所有空间

4、当第一次搜索到E位置时即可得到答案

核心：在每次三维延展时，第一次访问到的路径以后都不在访问，每条路径不相交，即可获得做最短路径！

代码：

#include <iostream>
#include <string.h>
using namespace std;
typedef class
{
	public:
	int l,r,c;
	int depth;	// 耗时 
} pf;
 
pf s,e;
bool maze[40][40][40];
int shortmin;

bool BFS(int i,int j,int k)
{
	bool vist[40][40][40]={false};
	pf queue [30000];
	int head,tail;
	queue[head=0].l=i;
	queue[tail=0].r=j;
	queue[0].c=k;
	queue[tail++].depth=0;
	
	vist[i][j][k]=true;
	
	while(head<tail)
	{
		pf x=queue[head++];
		
		if(x.l==e.l&&x.r==e.r&&x.c==e.c)
		{
			shortmin=x.depth;
			return true;
		}
		
		if(maze[x.l-1][x.r][x.c]&&!vist[x.l-1][x.r][x.c])
		{
			vist[x.l-1][x.r][x.c]=true;
			queue[tail].l=x.l-1;
			queue[tail].r=x.r;
			queue[tail].c=x.c;
			queue[tail++].depth=x.depth+1;
		}
		
		if(maze[x.l][x.r-1][x.c]&&!vist[x.l][x.r-1][x.c])
		{
			vist[x.l][x.r-1][x.c]=true;
			queue[tail].l=x.l;
			queue[tail].r=x.r-1;
			queue[tail].c=x.c;
			queue[tail++].depth=x.depth+1;
		}
		
		if(maze[x.l][x.r][x.c-1]&&!vist[x.l][x.r][x.c-1])
		{
			vist[x.l][x.r][x.c-1]=true;
			queue[tail].l=x.l;
			queue[tail].r=x.r;
			queue[tail].c=x.c-1;
			queue[tail++].depth=x.depth+1;
		}
		
		if(maze[x.l+1][x.r][x.c]&&!vist[x.l+1][x.r][x.c])
		{
			vist[x.l+1][x.r][x.c]=true;
			queue[tail].l=x.l+1;
			queue[tail].r=x.r;
			queue[tail].c=x.c;
			queue[tail++].depth=x.depth+1;
		}
		
		if(maze[x.l][x.r+1][x.c]&&!vist[x.l][x.r+1][x.c])
		{
			vist[x.l][x.r+1][x.c]=true;
			queue[tail].l=x.l;
			queue[tail].r=x.r+1;
			queue[tail].c=x.c;
			queue[tail++].depth=x.depth+1;
		}
		
		if(maze[x.l][x.r][x.c+1]&&!vist[x.l][x.r][x.c+1])
		{
			vist[x.l][x.r][x.c+1]=true;
			queue[tail].l=x.l;
			queue[tail].r=x.r;
			queue[tail].c=x.c+1;
			queue[tail++].depth=x.depth+1;
		}
	}	
	return false;
}

int main()
{
	std::ios::sync_with_stdio(false);
	int L,R,C;
	while(cin>>L>>R>>C&&L&&R&&C)
	{
		memset(maze,false,sizeof(maze));	//更新迷宫
		for(int i=1;i<=L;i++)
			for(int j=1;j<=R;j++)
				for(int k=1;k<=C;k++)
				{
					char temp;
					cin>>temp;
					if(temp=='.') maze[i][j][k]=true;
					if(temp=='S') 
					{
						maze[i][j][k]=true;
						s.l=i;
						s.r=j;
						s.c=k;
					}
					if(temp=='E') 
					{
						maze[i][j][k]=true;
						e.l=i;
						e.r=j;
						e.c=k;
					}
				 } 
		
		if(BFS(s.l,s.r,s.c))
		cout<<"Escaped in "<<shortmin<<" minute(s)."<<endl;
		else cout<<"Trapped!"<<endl;	
	 } 
	return 0;
}