BFS POJ2251 Dungeon Master
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
Is an escape possible? If yes, how long will it take?
Input
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
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
- POJ2251 Dungeon Master —— BFS
题目链接:http://poj.org/problem?id=2251 Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total S ...
- 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 ...
- 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 ...
- 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& ...
- bfs—Dungeon Master—poj2251
Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 32228 Accepted: 12378 ...
- 【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! ...
- POJ 2251 Dungeon Master(3D迷宫 bfs)
传送门 Dungeon Master Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 28416 Accepted: 11 ...
- 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 ...
- 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
- JMS基础篇
首先我们需要下载 ActiveMQ:http://activemq.apache.org/. 启动 ActiveMQ 服务:解包下载的 ActiveMQ >进去其bin 目录>双击 act ...
- BZOJ 2141: 排队 [CDQ分治]
题意: 交换序列中两个元素,求逆序对 做分块做到这道题...一看不是三维偏序嘛.... 作为不会树套树的蒟蒻就写CDQ分治吧.... 对时间分治...x排序...y树状数组... 交换拆成两个插入两个 ...
- BZOJ 3173: [Tjoi2013]最长上升子序列 [splay DP]
3173: [Tjoi2013]最长上升子序列 Time Limit: 10 Sec Memory Limit: 128 MBSubmit: 1613 Solved: 839[Submit][St ...
- Angular4---部署---Angular 与 Nginx的邂逅
Nginx + Angular结合操作 1.下载Nginx , 根据自己的版本下载Nginx,关于Nginx配置,请看https://www.cnblogs.com/MBirds/p/6605366. ...
- 说说VNode节点(Vue.js实现)
写在前面 因为对Vue.js很感兴趣,而且平时工作的技术栈也是Vue.js,这几个月花了些时间研究学习了一下Vue.js源码,并做了总结与输出.文章的原地址:https://github.com/an ...
- Windows 桌面和文件夹的右键->打开命令行窗口
Windows 桌面和文件夹的右键->打开命令行窗口 1.先按下shift,再点鼠标右键运行CMD,(不是管理员权限) 上图是我已经加了右键的,并且 系统设置了 ps代替cmd,所以是“在此处 ...
- 加入GIMPS项目,寻找梅森素数!
截止到目前为止人类共找到了50个梅森素数,其中最后16个梅森素数都是通过GIMPS项目找到的. 为了激励人们寻找梅森素数和促进网格技术发展,总部设在美国旧金山的电子前沿基金会(EFF)于1999年3月 ...
- pycharm的用法
Ctrl / 注释(取消注释)选择的行 Shift + Enter开始新行Ctrl + Enter智能换行TAB Shift+TAB缩进/取消缩进所选择的行Ctrl + Alt + I自动缩进行Ctr ...
- phpstorm及webstorm密钥
选用 server 方式,输入地址:http://idea.iteblog.com/key.php http://idea.lanyus.com/
- 机器学习之支持向量机(三):核函数和KKT条件的理解
注:关于支持向量机系列文章是借鉴大神的神作,加以自己的理解写成的:若对原作者有损请告知,我会及时处理.转载请标明来源. 序: 我在支持向量机系列中主要讲支持向量机的公式推导,第一部分讲到推出拉格朗日对 ...