Dungeon Master

Time Limit: 1000MS  Memory Limit: 10000K

Total Submissions: 40872  Accepted: 19936

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

 
三维的BFS。
 //2017-02-20
#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue> using namespace std; int dx[] = {, , , , -, };
int dy[] = {, , , -, , };
int dz[] = {, , , , , -}; struct node
{
int x, y, z, step;
}; int main()
{
int level, n, m;
char dun[][][];
bool vis[][][];
node tmp;
while(cin>>level>>n>>m)
{
if(level == && n == && m == )break;
queue<node> q;
memset(vis, , sizeof(vis));
for(int k = ; k < level; k++)
for(int i = ; i < n; i++)
for(int j = ; j < m; j++){
cin>>dun[i][j][k];
if(dun[i][j][k] == 'S'){
tmp.x = i;
tmp.y = j;
tmp.z = k;
tmp.step = ;
q.push(tmp);
vis[i][j][k] = ;
}
}
int x, y, z, step, nx, ny, nz;
bool ok = false;
while(!q.empty())
{
x = q.front().x;
y = q.front().y;
z = q.front().z;
step = q.front().step;
for(int i = ; i < ; i++)
{
nx = x+dx[i];
ny = y+dy[i];
nz = z+dz[i];
if(nx>=&&nx<n&&ny>=&&ny<m&&nz>=&&nz<level&&!vis[nx][ny][nz])
{
if(dun[nx][ny][nz] == '.'){
vis[nx][ny][nz] = ;
tmp.x = nx;
tmp.y = ny;
tmp.z = nz;
tmp.step = step+;
q.push(tmp);
}else if(dun[nx][ny][nz] == 'E'){
ok = true;
cout<<"Escaped in "<<step+<<" minute(s)."<<endl;
break;
}
}
}
if(ok)break;
q.pop();
}
if(!ok)cout<<"Trapped!"<<endl;
} return ;
}

POJ2251(KB1-B 三维BFS)的更多相关文章

  1. hdu 1240:Asteroids!(三维BFS搜索)

    Asteroids! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total ...

  2. 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 ...

  3. POJ 2049— Finding Nemo(三维BFS)10/200

    版权声明:本文为博主原创文章,未经博主同意不得转载. https://blog.csdn.net/u013497151/article/details/29562915 海底总动员.... 这个题開始 ...

  4. 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 ...

  5. AOJ.866 飞越原野 (三维BFS)

    AOJ.866 飞越原野 (三维BFS) 题意分析 点我挑战题目 相比于普通的BFS,要多一维来记录当前剩余的体力.而且还要额外的一层循环来处理,飞过的路程. 代码总览 #include <io ...

  6. SDUT OJ 1124 飞越原野 (三维BFS练习)

    飞跃原野 nid=24#time" title="C.C++.go.haskell.lua.pascal Time Limit5000ms Memory Limit 65536K ...

  7. Dungeon Master POJ-2251 三维BFS

    题目链接:http://poj.org/problem?id=2251 题目大意 你被困在了一个三维的迷宫,找出能通往出口的最短时间.如果走不到出口,输出被困. 思路 由于要找最短路径,其实就是BFS ...

  8. POJ2251——Dungeon Master(三维BFS)

    和迷宫问题区别不大,相比于POJ1321的棋盘问题,这里的BFS是三维的,即从4个方向变为6个方向. 用上队列的进出操作较为轻松. #include<iostream> #include& ...

  9. POJ-2251.DungeonMaster(三维BFS)

    做题时需要注意,爬楼有向上和向下爬之分... 本题大意:输入 l, r, c, 分别代表地牢的楼层数和每层地牢的长和宽,地牢由rock and point and source and key组成,你 ...

随机推荐

  1. 五,mysql优化——sql语句优化小技巧

    1,大批量插入数据 (1)对于MyISAM: alter table table_name disable keys; loading data; alter table table_name ena ...

  2. Android安全防护防护———Android 端常见的安全问题

    Android安全防护防护——加密算法:传送门https://www.cnblogs.com/huangjialin/p/9694488.html 组件安全 activity劫持 简单来说就是正常的a ...

  3. 关于使用Iscroll.js异步加载数据后不能滑动到最底端的问题解决方案

    关于使用Iscroll.js异步加载数据后不能滑动到最底端,拉到最下边又弹回去的问题困扰了我老半天,相信很多朋友都遇到了.我刚好不小心解决了,和大家分享一下.由于各种忙,下边就直接上代码吧. (前提是 ...

  4. Redis 的 Sentinel

    Redis 的 Sentinel 系统用于管理多个 Redis 服务器(instance), 该系统执行以下三个任务: 监控(Monitoring): Sentinel 会不断地检查你的主服务器和从服 ...

  5. LINUX中如何查看某个端口是否被占用

    之前查询端口是否被占用一直搞不明白,问了好多人,终于搞懂了,现在总结下: 1.netstat  -anp  |grep   端口号 如下,我以3306为例,netstat  -anp  |grep   ...

  6. python相见恨晚的库

    1)基本工具: virtualenv(虚拟环境)pip.setuptools (e.g. easy_install,这些东西肯定要呢)ipython(用了以后,就不再想用普通的python shell ...

  7. [Noi2014]购票 斜率优化DP+可持久化凸包

    貌似网上大部分题解都是CDQ分治+点分治然后再斜率优化DP,我貌似并没有用这个方法. 这一题跟这题有点像,只不过多了一个l的限制 如果说直接跑斜率优化DP,存储整个序列的话,显然是不行的,如图所示(图 ...

  8. js delete删除对象属性,delete删除不了变量及原型链中的变量

    js delete删除对象属性,delete删除不了变量及原型链中的变量 一.delete删除对象属性 function fun(){ this.name = 'gg'; } var obj = ne ...

  9. node开发环境配置

    node开发环境配置 用处 NodeJS——后台 JavaScript-前台 后台其他语言 1.PHP 2.Java 3.Pythonnode优势 1.性能高 nodejs php 86 1s 1分半 ...

  10. Oracle 数据库维护管理之--dbms_lock

    1.查询相关的v$视图,但是提示表或视图不存在解决办法     原因是使用的用户没有相关的查询权限导致 解决办法: grant select  any dictionary to 用户;    --这 ...