Robot Motion

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
There will be one or more grids for robots to navigate. The data for each is in the following form. On the first line are three integers separated by blanks: the number of rows in the grid, the number of columns in the grid, and 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
For each grid in the input there is one line of output. Either the robot follows a certain number of instructions and exits the grid on any one the four sides or else the robot follows the instructions on a certain number of locations 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)

题目大意:看图+样例基本就懂了。
解题思路:模拟
Code:

 #include<string>
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int a,b,first,px,py,date1,date2;
struct Point
{
char dir;
int vis,step;
} P[];
int move(int cnt)
{
int tmp=(py-)*a+px;
if (P[tmp].dir=='N') py--;
if (P[tmp].dir=='E') px++;
if (P[tmp].dir=='S') py++;
if (P[tmp].dir=='W') px--;
if (px>=a+||px<=||py>=b+||py<=)
return ;
tmp=(py-)*a+px;
if (P[tmp].vis)
{
date1=P[tmp].step-;
date2=cnt-date1-;
return -;
}
else P[tmp].vis=,P[tmp].step=cnt;
return ;
}
int main()
{
while (cin>>b>>a>>first)
{
int k=;
if (!a&&!b&&!first) break;
for (int i=; i<=b; i++)
for (int j=; j<=a; j++)
{
cin>>P[k].dir;
P[k++].vis=;
}
px=first,py=;
P[first].step=,P[first].vis=;
int cnt=,ok;
while ()
{
ok=move(cnt);
if (ok) break;
cnt++;
}
if (ok==) printf("%d step(s) to exit\n",cnt-);
else printf("%d step(s) before a loop of %d step(s)\n",date1,date2);
}
return ;
}

POJ1573——Robot Motion的更多相关文章

  1. poj1573 Robot Motion

    Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12507   Accepted: 6070 Des ...

  2. POJ-1573 Robot Motion模拟

    题目链接: https://vjudge.net/problem/POJ-1573 题目大意: 有一个N*M的区域,机器人从第一行的第几列进入,该区域全部由'N' , 'S' , 'W' , 'E' ...

  3. poj1573 Robot Motion(DFS)

    题目链接 http://poj.org/problem?id=1573 题意 一个机器人在给定的迷宫中行走,在迷宫中的特定位置只能按照特定的方向行走,有两种情况:①机器人按照方向序列走出迷宫,这时输出 ...

  4. POJ1573(Robot Motion)--简单模拟+简单dfs

    题目在这里 题意 : 问你按照图中所给的提示走,多少步能走出来??? 其实只要根据这个提示走下去就行了.模拟每一步就OK,因为下一步的操作和上一步一样,所以简单dfs.如果出现loop状态,只要记忆每 ...

  5. POJ1573 Robot Motion(模拟)

    题目链接. 分析: 很简单的一道题, #include <iostream> #include <cstring> #include <cstdio> #inclu ...

  6. 【POJ - 1573】Robot Motion

    -->Robot Motion 直接中文 Descriptions: 样例1 样例2 有一个N*M的区域,机器人从第一行的第几列进入,该区域全部由'N' , 'S' , 'W' , 'E' ,走 ...

  7. Robot Motion(imitate)

    Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11065   Accepted: 5378 Des ...

  8. 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 ...

  9. POJ 1573 Robot Motion(BFS)

    Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12856   Accepted: 6240 Des ...

随机推荐

  1. 【转帖】error C2296: “^”: 非法,左操作数包含“double”类型

    想要实现 ,写的C++程序为 double x; x=2^3; 结果程序总是出现这样的错误:error C2296: “^”: 非法,左操作数包含“double”类型 后来才发现操作符“^”,在C++ ...

  2. js设备判断

    判断设备android,weixin,ios var UserAgent=navigator.userAgent.toLowerCase(); var IS_ANDROID=false; var IS ...

  3. 什么是MBR?(含图解)

    Mbr位于磁盘的0柱面,0磁头,1扇区. MBR       有三部分构成,主引导程序,硬盘分区表DPT和,硬盘的有效标志55AA.在512个字节的主引导扇区里. 主引导程序占446个字节,dpt占6 ...

  4. tomcat6.0添加ssi(*.shtml)配置

    1.去掉tomcat6中conf/web.xml关于ssi的注释 <servlet> <servlet-name>ssi</servlet-name> <se ...

  5. linux 输入子系统(2)----简单实例分析系统结构(input_dev层)

    实例代码如下: #include <linux/input.h> #include <linux/module.h> #include <linux/init.h> ...

  6. opencv学习笔记(02)——遍历图像(指针法)

    #include <opencv2\core\core.hpp> #include <opencv2\highgui\highgui.hpp> #include <ope ...

  7. Kinetic使用注意点--ellipse

    new Ellipse(config) 参数: config:包含所有配置项的对象. { radius: "半径,可以用数字a.数组[a,b]或对象{x:a,y:b}来表示" } ...

  8. 【Cardboard】 体验 - Google Cardboard DIY及完成后简单体验

    体验 - Google Cardboard DIY及完成后简单体验 今年的Google I/O最让我感兴趣的除了Material Design以外就是这个Google Cardboard了.据说是Go ...

  9. 【BZOJ 1927】 [Sdoi2010]星际竞速

    Description 10 年一度的银河系赛车大赛又要开始了.作为全银河最盛大的活动之一, 夺得这个项目的冠军无疑是很多人的梦想,来自杰森座 α星的悠悠也是其中之一. 赛车大赛的赛场由 N 颗行星和 ...

  10. 对日期和时间的处理 NSCalendar

    代码较老,供参考 NSCalendar用于处理时间相关问题.比如比较时间前后.计算日期所的周别等. 1. 创建或初始化可用以下方法 + (id)currentCalendar; 取得当前用户的逻辑日历 ...