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. 系统架构之负载均衡【F5\nginx\LVS\DNS轮询\】

    在做系统架构规划的时候,负载均衡,HA(高可用性集群,是保证业务连续性的有效解决方案,一般有两个或两个以上的节点,且分为活动节点及备用节点,当活动节点出现故障的时候,由备用节点接管)都是经常需要考虑的 ...

  2. BZOJ4840 NEERC2016 Binary Code

    Problem BZOJ Solution 可能是因为快要省选了,所以最近更博的频率好像高了点_(:зゝ∠)_ 每个字符串最多有两个状态,然后要满足一些依赖关系,考虑2sat.可以先把字符串的结束节点 ...

  3. idea中使用tomcat 方式启动spring boot项目

    Spring boot 的main 入口启动方式相信都会用,直接运行main直接就启动了,但是往往这种方式并不是最佳的启动方式,比如运维的层面更希望调整tomcat的调优参数,而只使用嵌入启动方式很难 ...

  4. strcpy unsigned char

    http://bbs.csdn.net/topics/250068243 char *strcpy(char* dest, const char *src); 用unsigned char编译会出错 ...

  5. 缓存数据库-redis数据类型和操作(set)

    一:Redis 集合(Set) Redis的Set是string类型的无序集合.集合成员是唯一的,这就意味着集合中不能出现重复的数据. Redis 中 集合是通过哈希表实现的,所以添加,删除,查找的复 ...

  6. irport报表,把数字金额转换成大写人民币金额

    1.编写oracle函数 CREATE OR REPLACE Function MoneyToChinese(Money In Number) Return Varchar2 Is strYuan ) ...

  7. 在Eclipse中建立Maven工程

  8. git —— 分支

    git中每一个分支相当于一个时间线 并列且相互平行 控制用指针控制~ 1.第一种创建命令: $ git branch 分支名称 —— 创建分支 $ git checkout 分支名称 —— 切换分支 ...

  9. WordPress“无法将上传的文件移动至wp-content/uploads/”的解决办法

    WordPress“无法将上传的文件移动至wp-content/uploads/”的问题在有些配置不完善的虚拟主机或服务器中会碰到,一般会出现以下症状:1.无论是从文章编辑页面还是媒体库页面都无法上传 ...

  10. 关于Eclipse连接sql server 2008的若干问题

    以下内容转自:https://www.cnblogs.com/skylarzhan/p/7619977.html Eclipse中使用SQL server 2008数据库 一.准备材料 要能够使用数据 ...