Dungeon Master
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 17555   Accepted: 6835

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

 
bfs 六个方向直接乱搞就好
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 31
const int inf=0x7fffffff; //无限大
int visited[maxn][maxn][maxn];
struct node
{
int x;
int y;
int z;
int cnt;
};
int dx[]={,-,,,,};
int dy[]={,,,-,,};
int dz[]={,,,,,-};
int main()
{
int n,m,k;
string s;
while(cin>>n>>m>>k)
{
if(n==&&m==&&k==)
break;
memset(visited,,sizeof(visited));
node sb;
node sb2;
for(int i=;i<n;i++)
{
for(int j=;j<m;j++)
{
cin>>s;
for(int t=;t<k;t++)
{
if(s[t]=='S')
{
sb.x=i;
sb.y=j;
sb.z=t;
sb.cnt=;
}
if(s[t]=='E')
{
sb2.x=i;
sb2.y=j;
sb2.z=t;
}
if(s[t]=='#')
{
visited[i][j][t]=;
}
}
}
}
queue<node> q;
q.push(sb);
visited[sb.x][sb.y][sb.z]=;
int flag=;
while(!q.empty())
{
node now=q.front();
for(int i=;i<;i++)
{
node next;
next.x=now.x+dx[i];
next.y=now.y+dy[i];
next.z=now.z+dz[i];
next.cnt=now.cnt+;
if(next.x<||next.x>=n||next.y<||next.y>=m||next.z<||next.z>=k)
continue;
if(visited[next.x][next.y][next.z])
continue;
visited[next.x][next.y][next.z]=;
if(next.x==sb2.x&&next.y==sb2.y&&next.z==sb2.z)
flag=next.cnt;
if(flag>)
break;
q.push(next);
}
if(flag>)
break;
q.pop();
}
if(flag>)
printf("Escaped in %d minute(s).\n",flag);
else
cout<<"Trapped!"<<endl; }
return ;
}

hdu 2251 Dungeon Master bfs的更多相关文章

  1. poj 2251 Dungeon Master( bfs )

    题目:http://poj.org/problem?id=2251 简单三维 bfs不解释, 1A,     上代码 #include <iostream> #include<cst ...

  2. poj 2251 Dungeon Master (BFS 三维)

    You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of un ...

  3. POJ 2251 Dungeon Master bfs 难度:0

    http://poj.org/problem?id=2251 bfs,把两维换成三维,但是30*30*30=9e3的空间时间复杂度仍然足以承受 #include <cstdio> #inc ...

  4. POJ 2251 Dungeon Master (BFS最短路)

    三维空间里BFS最短路 #include <iostream> #include <cstdio> #include <cstring> #include < ...

  5. POJ 2251 Dungeon Master --- 三维BFS(用BFS求最短路)

    POJ 2251 题目大意: 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.不同L层 ...

  6. POJ 2251 Dungeon Master (三维BFS)

    题目链接:http://poj.org/problem?id=2251 Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total S ...

  7. POJ.2251 Dungeon Master (三维BFS)

    POJ.2251 Dungeon Master (三维BFS) 题意分析 你被困在一个3D地牢中且继续寻找最短路径逃生.地牢由立方体单位构成,立方体中不定会充满岩石.向上下前后左右移动一个单位需要一分 ...

  8. BFS POJ 2251 Dungeon Master

    题目传送门 /* BFS:这题很有意思,像是地下城,图是立体的,可以从上张图到下一张图的对应位置,那么也就是三维搜索,多了z坐标轴 */ #include <cstdio> #includ ...

  9. poj 2251 Dungeon Master

    http://poj.org/problem?id=2251 Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submis ...

随机推荐

  1. int、long、long long取值范围

    unsigned   int   0-4294967295 int   -2147483648-2147483647 unsigned long 0-4294967295 long   -214748 ...

  2. Python生成器-博文读后感

    Windows 10家庭中文版,Python 3.6.4, 上午看过了一篇讲Python生成器的博文: 提高你的Python: 解释‘yield’和‘Generators(生成器)’(英文原文) 这篇 ...

  3. java基础81 jsp的内置对象(网页知识)

    1.什么是内置对象? 在jsp开发中,会频繁使用到一些对象,如:HttpSession,ServletContext,HttpServletRequest.      如果每次使用这些对象时,都要去创 ...

  4. supervisor的安装和配置

    1. 安装 yum install supervisor 2.配置 [unix_http_server] file=/tmp/supervisor.sock ;UNIX socket 文件,super ...

  5. tp杂记

    /** php中的大U函数三个参数: U('ajaxDelPic') ==> /index.php/Admin/Goods/ajaxDelPic.html U('ajaxDelPic?id=1' ...

  6. springMVC源码分析--HttpMessageConverter参数read操作(二)

    上一篇博客springMVC源码分析--HttpMessageConverter数据转化(一)中我们简单介绍了一下HttpMessageConverter接口提供的几个方法,主要有以下几个方法: (1 ...

  7. 13 在 O(1) 时间内删除链表节点

    删除链表的一个结点,用下一个结点覆盖掉要删除的结点,再释放掉要删结点的下一个结点的内存 Java: public ListNode deleteNode(ListNode head, ListNode ...

  8. Spark(十五)SparkCore的源码解读

    一.启动脚本分析 独立部署模式下,主要由master和slaves组成,master可以利用zk实现高可用性,其driver,work,app等信息可以持久化到zk上:slaves由一台至多台主机构成 ...

  9. codeforces 354 D. Transferring Pyramid

    D. Transferring Pyramid time limit per test 3 seconds memory limit per test 256 megabytes input stan ...

  10. 收集Nginx的json格式日志(五)

    一.配置nginx [root@linux-node1 ~]# vim /etc/nginx/nginx.conf #修改日志格式为json格式,并创建一个nginxweb的网站目录 log_form ...