NOI2.5 1253:Dungeon Master

描述

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?

输入

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.

输出

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!

 样例输入3 4 5
S. . . .
. ###.
. ##. .
###.#

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

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

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

0 0 0
样例输出Escaped in 11 minute(s).
Trapped!
 

题目大意：

给定一个三维迷宫，给出起点（S）和终点（T），问从起点到终点要走几分钟（走一步一分钟），如果走不到终点，输出“Trapped!”

三维？！你提莫的在逗我？！二维已经够呛了，你给我来三维！！！

细想一下，也不是很难，广搜的话，多加两个方向的走法就可以了呀

队列再加一个来保存层数就行了呀

伪队列写法，一次AC

 

代码如下：

<span style="font-size:12px;BACKGROUND-COLOR: #ffff99">#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
using namespace std;
char c[35];
int f[6]={1,-1,0,0,0,0},u[6]={0,0,-1,1,0,0},g[6]={0,0,0,0,-1,1},l,m,n,xz,yz,zz,num,nn;
bool v[35][35][35];
void find(int p,int q,int r)
{
	int x=0,y=0,z=0,t=0,w=1,i;
	int h[100001][4];
	h[1][0]=p;
	h[1][1]=q;
	h[1][2]=r;
	h[1][3]=0;
	do
	{
		t++;
		for(i=0;i<6;i++)
		{
			x=h[t][0]+f[i];
			y=h[t][1]+u[i];
			z=h[t][2]+g[i];
			if(x>=0&&x<m&&y>=0&&y<n&&z>=0&&z<l&&!v[x][y][z])
			{
				w++;
				h[w][0]=x;
				h[w][1]=y;
				h[w][2]=z;
				h[w][3]=h[t][3]+1;
				v[x][y][z]=1;
				if(x==xz&&y==yz&&z==zz)
				{
					num=h[w][3];
					return ;
				}
			}
		}
	}while(t<w);
}
int main()
{
	int i,j,x,y,z;
	scanf("%d%d%d",&l,&m,&n);
	while(l&&m&&n)
	{
		num=0;
		for(int o=0;o<l;o++)
		for(i=0;i<m;i++)
		{
			scanf("%s",c);
			for(j=0;j<=n;j++)
			{
				if(c[j]=='#')
					v[i][j][o]=1;
				if(c[j]=='S')
				{
					x=i;
					y=j;
					z=o;
					v[i][j][o]=0;
				}
				if(c[j]=='E')
				{
					xz=i;
					yz=j;
					zz=o;
					v[i][j][o]=0;
				}
				if(c[j]=='.')
					v[i][j][o]=0;
			}
		}
		v[x][y][z]=1;
		find(x,y,z);
		if(num)
			printf("Escaped in %d minute(s).
",num);
		else
			printf("Trapped!
");
		scanf("%d%d%d",&l,&m,&n);
	}
}</span>

<span style="font-size:12px;BACKGROUND-COLOR: #ffff99"></span> 

伪队列的方法很节约时间，即使三维也不会很耗时，所以，就这样啦