B - Dungeon Master

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

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!

BFS水题,套模板按6个方向拓展就行了。

 #include <iostream>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <string.h>
using namespace std;
int to[][] = {{,,},{,,-},{,,},{,-,},{,,},{-,,}};
char s[][][];
int maps[][][];
int x1,y1,z1;
int x2,y2,z2;
int L,R,C;
struct data
{
int x;
int y;
int z;
int step;
};
int check(int x,int y,int z)
{
if(x< || y< || z< || x>=L || y>=R || z>=C)
return ;
else if(s[x][y][z] == '#')
return ;
else if(maps[x][y][z])
return ;
return ;
}
int bfs()
{
queue<data> q;
data now,nex;
now.x=x1;
now.y=y1;
now.z=z1;
now.step=;
maps[x1][y1][z1]=;
q.push(now);
while(!q.empty())
{
now=q.front();
q.pop();
if(now.x==x2&&now.y==y2&&now.z==z2)
{
return now.step;
}
for(int i=;i<;i++)
{ nex=now;
nex.x=now.x+to[i][];
nex.y=now.y+to[i][];
nex.z=now.z+to[i][];
if(check(nex.x,nex.y,nex.z))
{
continue;
}
maps[nex.x][nex.y][nex.z]=;
nex.step=now.step+;
q.push(nex);
}
}
return ;
}
int main()
{
while(scanf("%d%d%d",&L,&R,&C),L+R+C)
{
cin>>R>>C;
int i,j,k;
for(i=;i<L;i++) //
{
for(j=;j<R;j++)
{
scanf("%s",s[i][j]);
for(k=;k<C;k++)
{
if(s[i][j][k]=='S')
{
x1=i;
y1=j;
z1=k;
}
else if(s[i][j][k]=='E')
{
x2=i;
y2=j;
z2=k;
}
}
}
}
memset(maps,,sizeof(maps));
int ans;
ans=bfs();
if(ans)
{printf("Escaped in %d minute(s).\n",ans);}
else
{
printf("Trapped!\n");
}
}
return ;
}

BFS POJ2251 Dungeon Master的更多相关文章

  1. POJ2251 Dungeon Master —— BFS

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

  2. POJ-2251 Dungeon Master (BFS模板题)

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

  3. POJ2251 Dungeon Master(bfs)

    题目链接. 题目大意: 三维迷宫,搜索从s到e的最小步骤数. 分析: #include <iostream> #include <cstdio> #include <cs ...

  4. 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& ...

  5. bfs—Dungeon Master—poj2251

    Dungeon Master Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 32228   Accepted: 12378 ...

  6. 【POJ - 2251】Dungeon Master (bfs+优先队列)

    Dungeon Master  Descriptions: You are trapped in a 3D dungeon and need to find the quickest way out! ...

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

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

  8. 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 ...

  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. Spring源码情操陶冶-PathMatchingResourcePatternResolver路径资源匹配溶解器

    本文简单的分析下spring对某个目录下的class资源是如何做到全部的加载 PathMatchingResourcePatternResolver#getResources PathMatching ...

  2. Using Custom Domains With IIS Express In Asp.Net Core

    IIS Express是一个Mini版的IIS,能够支持所有的Web开发任务,但是这种设计有一些缺陷,例如只能通过localhost:<port>的方式来访问我们的应用程序,看起来就有点不 ...

  3. k短路模板 POJ2449

    采用A*算法的k短路模板 #include <iostream> #include <cstdio> #include <cstring> #include < ...

  4. BZOJ 3994: [SDOI2015]约数个数和 [莫比乌斯反演 转化]

    2015 题意:\(d(i)\)为i的约数个数,求\(\sum\limits_{i=1}^n \sum\limits_{j=1}^m d(ij)\) \(ij\)都爆int了.... 一开始想容斥一下 ...

  5. BZOJ 2199: [Usaco2011 Jan]奶牛议会 [2-SAT 判断解]

    http://www.lydsy.com/JudgeOnline/problem.php?id=2199 题意:裸的2-SAT,但是问每个变量在所有解中是只能为真还是只能为假还是既可以为真又可以为假 ...

  6. 在 CentOS7.0 上搭建 Chroot 的 Bind DNS 服务器

    BIND(Berkeley internet Name Daemon)也叫做NAMED,是现今互联网上使用最为广泛的DNS 服务器程序.这篇文章将要讲述如何在 chroot 监牢中运行 BIND,这样 ...

  7. 【NOIP2012】 疫情控制

    [NOIP2012] 疫情控制 标签: 倍增 贪心 二分答案 NOIP Description H 国有 n 个城市,这 n 个城市用 n-1 条双向道路相互连通构成一棵树, 1 号城市是首都, 也是 ...

  8. 卷积神经网络(CNN)在句子建模上的应用

    之前的博文已经介绍了CNN的基本原理,本文将大概总结一下最近CNN在NLP中的句子建模(或者句子表示)方面的应用情况,主要阅读了以下的文献: Kim Y. Convolutional neural n ...

  9. Json对象与Json字符串互转(4种转换方式)(转)

    1>jQuery插件支持的转换方式: $.parseJSON( jsonstr ); //jQuery.parseJSON(jsonstr),可以将json字符串转换成json对象  2> ...

  10. vim操作备忘录

    vim操作备忘录 vim 备忘录 vim的书籍虽然看不不少,可是老是容易忘记,主要是自己操作总结过少,这个博客就主要用来记录一些比较常见的术语和操作,以防止自己再次忘记. <leader> ...