hdu-1573 Robot Motion
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 10219 | Accepted: 4977 |
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
the number of the column in which the robot enters from the north. The possible entry columns are numbered starting with one at the left. Then come the rows of the direction instructions. Each grid will have at least one and at most 10 rows and columns of
instructions. The lines of instructions contain only the characters N, S, E, or W with no blanks. The end of input is indicated by a row containing 0 0 0.
Output
once, and then the instructions on some number of locations repeatedly. The sample input below corresponds to the two grids above and illustrates the two forms of output. The word "step" is always immediately followed by "(s)" whether or not the number before
it is 1.
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
//主要是推断是否构成循环。依据题目特点。能够设置一个标记数组,假设走过了就不能在走,直到退出循环。
用另外一个数组表示到这一个坐标的步数。那么退出循环的时候事实上已经到曾经走过的点。输出步数,然后循环节就是总步数减去当前步数。
#include<stdio.h>
#include<string.h>
int main()
{
int n,m,k,l,p,step;
char s[11][11];
int map[11][11];
while(scanf("%d %d",&n,&m)!=EOF)
{
if(n==0||m==0)break;
scanf("%d",&k);
for(int i=0;i<n;i++)
scanf("%s",s[i]);
l=0;
p=k-1;
step=0;
while(l>=0&&l<n&&p>=0&&p<m&&s[l][p]!='0')
{
if(s[l][p]=='N')
{
s[l][p]='0';
map[l][p]=++step; //先计算当前步数,然后在改变坐标值
l--;
}
else if(s[l][p]=='S')
{
s[l][p]='0';
map[l][p]=++step;
l++;
}
else if(s[l][p]=='E')
{
s[l][p]='0';
map[l][p]=++step;
p++;
}
else if(s[l][p]=='W')
{
s[l][p]='0';
map[l][p]=++step;
p--;
}
}
if(s[l][p]=='0') //表示循环 了、
printf("%d step(s) before a loop of %d step(s)\n",map[l][p]-1,step+1-map[l][p]);
else printf("%d step(s) to exit\n",step);
}
return 0;
}
hdu-1573 Robot Motion的更多相关文章
- 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 ...
- HDOJ(HDU).1035 Robot Motion (DFS)
HDOJ(HDU).1035 Robot Motion [从零开始DFS(4)] 点我挑战题目 从零开始DFS HDOJ.1342 Lotto [从零开始DFS(0)] - DFS思想与框架/双重DF ...
- POJ 1573 Robot Motion(BFS)
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12856 Accepted: 6240 Des ...
- 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 ...
- POJ 1573 Robot Motion(模拟)
题目代号:POJ 1573 题目链接:http://poj.org/problem?id=1573 Language: Default Robot Motion Time Limit: 1000MS ...
- POJ 1573 Robot Motion
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12978 Accepted: 6290 Des ...
- poj 1573 Robot Motion【模拟题 写个while循环一直到机器人跳出来】
...
- [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. ...
- hdu 1035 Robot Motion(模拟)
Problem Description A robot has been programmed to follow the instructions in its path. Instructions ...
- hdu 1035 Robot Motion(dfs)
虽然做出来了,还是很失望的!!! 加油!!!还是慢慢来吧!!! >>>>>>>>>>>>>>>>> ...
随机推荐
- JavaScripts学习日记——DOM SAX JAXP DEMO4J XPath
今日关键词: XML解析器 DOM SAX JAXP DEMO4J XPath XML解析器 1.解析器概述 什么是解析器 XML是保存数据的文件,XML中保存的数据也需要被程序读取然后使用.那么程序 ...
- 网页CSS2
列表与方块 width , hight (top, bottom ,left , right) 只有在决对坐标下才起作用 下面的使用与 ol ul list-style:none // 取消序号 ...
- django: startproject
python 的 django 框架的安装教程很多,这里不列举安装过程,直接开始记开发应用过程. 1 startprojec,新建项目 $ django-admin.py startproject ...
- 调用有道翻译API
前两天朋友说起NASA开放了数据API,今儿突然想起从来没用过外部提供的API,然而简单用得多的貌似是有道词典API,就像试试,本来觉得应该挺简单的,用urllib模块很快就实现了. 不过测试时才发现 ...
- DataGrid 简单数据绑定实例1
1.默认数据显示(自动显示列) 后台绑定 //DataGrid 数据绑定 dataGridOne.ItemsSource = _Context.Info.ToList(); 前台定义 <Data ...
- 关于看似简单的eclipse中tomcat小猫图标消失的问题解决
首先,这个问题出现在我新安装的虚拟机中,自己准备重新搭一套开发环境用于学习. 所以,出于好奇,自己从官网上把eclipse的最新版neo下下来尝尝鲜,刚安装好后发现与之前用的旧版基本相同,于是把相应的 ...
- CentOS6.5 PHP基础环境搭建 [个人整理-亲测可用]
** * CentOS6.5 搭建基础PHP环境(yum安装) * http://www.aiplaypc.com/160.html ** #安装需要的包,有依赖关系,自动帮你解决 yum ins ...
- 带权并查集 poj1182
首先要注意核心代码 int find(int i){ if(i == fa[i]) return fa[i]; int tt = find(fa[i]); num[i] ...
- (转)函数中使用 ajax 异步 同步 返回值错误 主函数显示返回值总是undefined -- ajax使用总结
aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAAloAAAE0CAIAAAB7LwoKAAAgAElEQVR4nO2dy6sc152A6+/R2mXwSn ...
- js formatString 格式化字符串
/* 函数:格式化字符串 参数:str:字符串模板: data:数据 调用方式:formatString("api/values/{id}/{name}",{id:101,name ...