POJ 1573 Robot Motion(模拟)
题目代号:POJ 1573
题目链接:http://poj.org/problem?id=1573
Default
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 14195 | Accepted: 6827 |
Description

A robot has been programmed to follow the instructions in its path. Instructions for the next direction the robot is to move are laid down in a grid. The possible instructions are
N north (up the page)
S south (down the page)
E east (to the right on the page)
W west (to the left on the page)
For example, suppose the robot starts on the north (top) side of Grid 1 and starts south (down). The path the robot follows is shown. The robot goes through 10 instructions in the grid before leaving the grid.
Compare what happens in Grid 2: the robot goes through 3 instructions only once, and then starts a loop through 8 instructions, and never exits.
You are to write a program that determines how long it takes a robot to get out of the grid or how the robot loops around.
Input
Output
Sample Input
3 6 5
NEESWE
WWWESS
SNWWWW
4 5 1
SESWE
EESNW
NWEEN
EWSEN
0 0 0
Sample Output
10 step(s) to exit
3 step(s) before a loop of 8 step(s)
Source
题目大意:WESN分别代表四个方向,第三个数据代表第一行的第几个位置开始,如果能走出去则输出走了几步,如果不能走出去进入了循环则输出第几步进入了循环,循环有几步。按题目标准格式输出。
解题思路:水题,作个标记代表第几步,如果下一步被标记过了则退出循环输出,如果走出去了也退出循环。
AC 代码:
# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL; char a[][];
int b[][]; int main()
{
//freopen("in.txt", "r", stdin);
int n,m,k;
while(cin>>n>>m,n&&m)
{
cin>>k;
mem(a,);
mem(b,);
for(int i=;i<=n;i++)
cin>>a[i]+;
int x=,y=k;
b[x][y]=;
int ans=;
int flag=;
while()
{
if(a[x][y]=='W')y--;
else if(a[x][y]=='S')x++;
else if(a[x][y]=='E')y++;
else if(a[x][y]=='N')x--;
if(x==||x==n+||y==||y==m+)
{
printf("%d step(s) to exit\n",ans);
break;
}
else if(b[x][y])
{
printf("%d step(s) before a loop of %d step(s)\n",b[x][y]-,ans-b[x][y]+);
break;
}
else
{
b[x][y]=++ans;
}
}
}
return ;
}
POJ 1573 Robot Motion(模拟)的更多相关文章
- POJ 1573 Robot Motion 模拟 难度:0
#define ONLINE_JUDGE #include<cstdio> #include <cstring> #include <algorithm> usin ...
- 模拟 POJ 1573 Robot Motion
题目地址:http://poj.org/problem?id=1573 /* 题意:给定地图和起始位置,robot(上下左右)一步一步去走,问走出地图的步数 如果是死循环,输出走进死循环之前的步数和死 ...
- poj 1573 Robot Motion【模拟题 写个while循环一直到机器人跳出来】
...
- POJ 1573 Robot Motion(BFS)
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12856 Accepted: 6240 Des ...
- POJ 1573 Robot Motion
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12978 Accepted: 6290 Des ...
- poj 1573 Robot Motion_模拟
又是被自己的方向搞混了 题意:走出去和遇到之前走过的就输出. #include <cstdlib> #include <iostream> #include<cstdio ...
- Poj OpenJudge 百练 1573 Robot Motion
1.Link: http://poj.org/problem?id=1573 http://bailian.openjudge.cn/practice/1573/ 2.Content: Robot M ...
- [ACM] hdu 1035 Robot Motion (模拟或DFS)
Robot Motion Problem Description A robot has been programmed to follow the instructions in its path. ...
- PKU 1573 Robot Motion(简单模拟)
原题大意:原题链接 给出一个矩阵(矩阵中的元素均为方向英文字母),和人的初始位置,问是否能根据这些英文字母走出矩阵.(因为有可能形成环而走不出去) 此题虽然属于水题,但是完全独立完成而且直接1A还是很 ...
随机推荐
- ll按时间排序和查看目录下文件数
查询文件并以降序排列:ll -t 查询文件并以升序排列:ll -t | tac 查询目录下文件数:ll|wc -l
- Oracle常用启停命令
一.监听启停 Oracle监听的启动.停止和状态查看 Oracle监听启动: lsnrctl start Oracle监听停止: lsnrctl stop Oracle监听状态 lsnrctl sta ...
- PHP 模拟http 请求
php 模拟请求类 <?php /** * fangdasheng * http 模拟请求 */ class Myhttp { private $apiUrl; // 构造函数 public f ...
- maven配置生成可执行的jar:maven-shade-plugin
默认打包生成的jar是不能直接运行的,因为带有main方法的信息不会添加到mainifest中,需要借助maven-shade-plugin <project> ... <build ...
- visual studio git 忽略文件配置模板
## Ignore Visual Studio temporary files, build results, and## files generated by popular Visual Stud ...
- js 类型转变
在绝大部分情况下,操作符和函数可以自动将一个值转换成正确的数据类型.这被称为"类型转变(type conversion)". 举个例子,alert 自动转变任何类型的参数为字符串类 ...
- 六、while循环
案例1: do while 循环 很少用到. for循环和while循环用的最多.
- vue——echarts更换主题
链接:https://blog.csdn.net/Sunshine0508/article/details/90067437 //等配置安装好了以后 在main.js里引入echarts主题的js,一 ...
- puppet之资源
资源 资源的定义 一个帐号,一个文件,目录,软件包都可以称作是资源,每个资源的定义都具有标题,类型,以及一些列属性. 常见的资源有notify(调试与输出),file(配置文件),package(软件 ...
- CentOS7安装并使用Ceph
1.准备工作1.1 安装配置NTP官方建议在所有 Ceph 节点上安装 NTP 服务(特别是 Ceph Monitor 节点),以免因时钟漂移导致故障. ln -sf /usr/share/zonei ...