Dungeon Master
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 21242   Accepted: 8265

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!
三维BFS 注意xyz别弄混了
/* ***********************************************
Author :pk29
Created Time :2015/8/19 18:40:40
File Name :4.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << ;
const double eps=1e-;
using namespace std;
struct node{
int x,y,z;
int dist;
};
bool cmp(int a,int b){
return a>b;
}
int l,r,c;
char mp[][][];
int vis[][][];
int sx,sy,sz,ex,ey,ez;
int dir[][]={,,,,,,,,-,,,,-,,,,-,}; void bfs(){
node u,v;
u.x=sx,u.y=sy,u.z=sz, u.dist=;
queue<node>q;
q.push(u);
int mark=;
vis[u.x][u.y][u.z]=;
while(!q.empty()){
u=q.front(),q.pop();
if(u.x==ex&&u.y==ey&&u.z==ez){
mark=;printf("Escaped in %d minute(s).\n",u.dist);break;
}
for(int i=;i<;i++){
int nx=u.x+dir[i][];
int ny=u.y+dir[i][];
int nz=u.z+dir[i][];
if(!vis[nx][ny][nz]&&mp[nx][ny][nz]!='#'&&nx>=&&ny>=&&nz>=&&nx<=r&&ny<=c&&nz<=l){
v.x=nx,v.y=ny,v.z=nz;
v.dist=u.dist+;
q.push(v);
vis[nx][ny][nz]=;
}
}
}
if(!mark)printf("Trapped!\n");
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
//freopen("out.txt","w",stdout);
while(cin>>l>>r>>c){
if(l==&&c==&&r==)break;
for(int i=;i<=l;i++)
for(int j=;j<=r;j++)
for(int k=;k<=c;k++){
cin>>mp[j][k][i];
if(mp[j][k][i]=='S')sx=j,sy=k,sz=i;
if(mp[j][k][i]=='E')ex=j,ey=k,ez=i;
}
cle(vis);
//cout<<sx<<sy<<sz<<endl;
//cout<<ex<<ey<<ez<<endl; bfs();
}
return ;
}

POJ 3279 Dungeon Master的更多相关文章

  1. 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 ...

  2. POJ 2251 Dungeon Master(地牢大师)

    p.MsoNormal { margin-bottom: 10.0000pt; font-family: Tahoma; font-size: 11.0000pt } h1 { margin-top: ...

  3. POJ 2251 Dungeon Master /UVA 532 Dungeon Master / ZOJ 1940 Dungeon Master(广度优先搜索)

    POJ 2251 Dungeon Master /UVA 532 Dungeon Master / ZOJ 1940 Dungeon Master(广度优先搜索) Description You ar ...

  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. 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 ...

  6. poj 2251 Dungeon Master

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

  7. 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 ...

  8. POJ 2251 Dungeon Master(3D迷宫 bfs)

    传送门 Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 28416   Accepted: 11 ...

  9. POJ 2251 Dungeon Master (非三维bfs)

    Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 55224   Accepted: 20493 ...

随机推荐

  1. Classloader中loadClass()方法和Class.forName()区别

    Classloader中loadClass()方法和Class.forName()都能得到一个class对象,那这两者得到的class对象有什么区别呢 1.java类装载的过程 Java类装载有三个步 ...

  2. mysql查询死锁,执行语句,服务器状态等语句集合

    [转]http://blog.csdn.net/enweitech/article/details/52447006

  3. redmine与SVN的Https方式整合问题

    尼玛啊!这个SVN的整合搞了一晚上,今天早上终于搞定了,FUCK!!! 进入话题: 可以先在bitnami redmine stack的命令行环境下手工运行svn,看是否能取到数据, svn list ...

  4. CodeForces 762D Maximum path

    http://codeforces.com/problemset/problem/762/D 因为是3*n很巧妙的地方是 往左走两步或更多的走法都可以用往回走以一步 并走完一列来替换 那么走的方法就大 ...

  5. call 和 apply 方法区别

    在js中call和apply它们的作用都是将函数绑定到另外一个对象上去运行,两者仅在定义参数方式有所区别,下面我来给大家介绍一下call和apply用法. 在web前端开发过程中,我们经常需要改变th ...

  6. Codeforces 713C Sonya and Problem Wihtout a Legend(DP)

    题目链接   Sonya and Problem Wihtout a Legend 题意  给定一个长度为n的序列,你可以对每个元素进行$+1$或$-1$的操作,每次操作代价为$1$. 求把原序列变成 ...

  7. codevs——1501 二叉树最大宽度和高度

    1501 二叉树最大宽度和高度  时间限制: 1 s  空间限制: 128000 KB  题目等级 : 白银 Silver 题解       题目描述 Description 给出一个二叉树,输出它的 ...

  8. #ifdef #endif #if #endif

    c语言里所有以#开头的都是预编译指令,就是在正式编译之前,让编译器做一些预处理的工作. #ifdef DEBUG printf("variable x has value = %d\n&qu ...

  9. 如何稳定地使用 Google 搜索https://encrypted.google.com/

    方法很简单.用记事本打开 hosts 文件(Windows Vista 和 Windows 7 用户请先使用管理员权限打开记事本,然后将 hosts 文件拖进记事本中),在最下面添加如下内容: 203 ...

  10. 【lombok】使用lombok注解,在代码编写过程中可以调用到get/set方法,但是在编译的时候无法通过,提示找不到get/set方法

    错误如题:使用lombok注解,在代码编写过程中可以调用到get/set方法,但是在编译的时候无法通过,提示找不到get/set方法 报错如下: 解决方法: 1.首先查看你的lombok插件是否下载安 ...