1010 Robot Motion
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.
题意:题目给出一个矩阵,让机器人按照规定行走;如果最终走入一个循环中,则输出进入循环前的的步数和循环的步数,如果最终走出矩阵范围也是输行走的步数;
规则如下:
E:向右走一步;W:向左走一步;N:向上走一步;S向下走一步;
AC代码:
#include<iostream>
#include<cstring>
#include<cstdio> using namespace std; int dp[][]={};
char ch[][];
int x,y,s;
int em()
{
int number=;
int a=,b=s;
while(){
if(dp[a][b]){//循环情况的处理;
cout<<dp[a][b]-<<" step(s) before a loop of "<<number-dp[a][b]<<" step(s)"<<endl;
return ;
}
if(a==||a==x+||b==||b==y+){cout<<number-<<" step(s) to exit"<<endl;return ;}
dp[a][b]=number;
switch(ch[a][b])
{
case 'E':b++;break;
case 'W':b--;break;
case 'S':a++;break;
case 'N':a--;break;
}
number++;
}
} int main()
{
freopen("1.txt","r",stdin);
int i,j;
while(){
memset(dp,,sizeof(dp));
cin>>x>>y>>s;
if(x==y&&y==s&&s==)return ;
for(i=;i<=x;i++)
for(j=;j<=y;j++)cin>>ch[i][j];
em();
}
}
1010 Robot Motion的更多相关文章
- 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 ...
- hdu1035 Robot Motion (DFS)
Robot Motion Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Tot ...
- poj1573 Robot Motion
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12507 Accepted: 6070 Des ...
- Robot Motion(imitate)
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 11065 Accepted: 5378 Des ...
- 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(BFS)
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12856 Accepted: 6240 Des ...
- Robot Motion 分类: POJ 2015-06-29 13:45 11人阅读 评论(0) 收藏
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 11262 Accepted: 5482 Descrip ...
- POJ 1573 Robot Motion
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12978 Accepted: 6290 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 ...
随机推荐
- Java 内部类 this
内部类访问外部类的一个例子: public class Abc { private class Bc { public void print() { System.out.println(Abc.th ...
- WWW 资源下载与表单提交
在注册与验证用户信息,以及非即时通信的游戏中,我们可以使用WWW类使用短链接来完成客户端与服务器数据的通信,今天我们将使用用POST方法来完成的用户注册与登录,在最后介绍下其它资源的加载. 首先使用P ...
- Elasticsearch常用插件(三)
elasticsearch-head 一个elasticsearch的集群管理工具,它是完全由html5编写的独立网页程序,你可以通过插件把它集成到es. 项目地址:https://github.co ...
- 改造vim
1.安装Vim和Vim基本插件首先安装好Vim和Vim的基本插件.这些使用apt-get安装即可: lingd@ubuntu:~/arm$sudo apt-get install vim vim-sc ...
- Masonry 添加约束要注意顺序
对一个视图添加约束,其依赖的约束必须先已经存在,不能依赖该代码后的约束,否则造成不可预料的结果,如下代码能达到预期效果 - (void)makeConstraints { __weak typeof( ...
- 手动安装VS code 插件
现在安装包: 通过修改下面的地址参数:https://${publisher}.gallery.vsassets.io/_apis/public/gallery/publisher/${publish ...
- Git 添加自己分支 pull request
1.找到项目地址 这里,我们可以找到项目地址,比如:https://github.com/*****/Cplusplus_Thread_Lib,然后点击页面右上角的 "fork" ...
- Hbulider里面template模板自用
template.js 一款 JavaScript 模板引擎,简单,好用.提供一套模板语法,用户可以写一个模板区块,每次根据传入的数据,生成对应数据产生的HTML片段,渲染不同的效果. 特性: 模版编 ...
- drupal 连表查询+分页
$query = db_select('Table','t'); $query->join('Table_A','a','on条件); $query->join('Table_B','b' ...
- ® 不需要显示为商标符的做法
若url中有参数reg,则把® 变为 ®