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
- Android 属性动画(二)
当然了属性动画还有一部分的知识点,也能做出很不错的效果,将在本篇博客为您展示~ 1.如何使用xml文件来创建属性动画 大家肯定都清楚,View Animator .Drawable Animator都 ...
- Python 文件I/O (转)
Python 文件I/O 本章只讲述所有基本的的I/O函数,更多函数请参考Python标准文档. 打印到屏幕 最简单的输出方法是用print语句,你可以给它传递零个或多个用逗号隔开的表达式.此函数把你 ...
- 笔记本分享无线Wifi
两种方法: 一.使用软件分享,如Wifi共享精灵,设置非常简单. 二.开启windows 7的隐藏功能:虚拟WiFi和SoftAP(即虚拟无线AP),就可以让电脑变成无线路由器,实现共享上网,节省网费 ...
- NHibernate——基本映射(5)
一.映射定义概括 1.1 映射定义(Mapping declaration) 对象和关系数据库之间的映射是用一个XML文档(XML document)来定义的.这个映射文档被设计为易读的,并且可以手工 ...
- (转)ip地址,手机ip查询
页面地址:http://www.ip138.com/ 外链地址(实际主页面里面有)http://www.ip138.com/iplink.htm 外链地址里面的内容: <FORM METHOD= ...
- (转)Facade模式
Facade模式要求一个子系统的外部与其内部的通信必须通过一个统一的Facade对象进行.Facade模式提供一个高层次的接口,使得子系统更易于使用. 就如同医院的接待员一样,Facade模式的Fac ...
- Android-兼容问题
兼容性问题从何而来?答:使用了低版本没有的功能,可是目标设备包括低版本. 那我们如何解决兼容性问题? 1.如果是JAVA代码 1.1 使用TargetApi(x)注解,避免Android Lint报错 ...
- C#生成Code39(extend)条形码【非条形码字体】
Code39是条形码的一种.由于编制简单.能够对任意长度的数据进行编码.支持设备广泛等特性而被广泛采用. 能够对任意长度的数据进行编码.其局限在于印刷品的长度和条码阅读器的识别范围. 支持设备广泛.目 ...
- VS2013配合EgretVS开发简单塔防游戏
VS2013配合EgretVS开发简单塔防游戏(1) - 环境配置 VS2013配合EgretVS开发简单塔防游戏(2) – 原型设计 VS2013配合EgretVS开发简单塔防游戏(3) – 精灵动 ...
- sass转换为css
sass安装的方法参考官网:http://www.w3cplus.com/sassguide/ SASS文件转换为CSS文件的方法: 首先输出 F: 代表找到F盘 : 然后输出cd sass 代表找到 ...