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

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)

Source

 #include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int row , col , o ;
char map[][] ;
int a[][] ;
int x , y ; void loop (char dir)
{
switch (dir)
{
case 'N' : printf ("%d step(s) before a loop of %d step(s)\n" , a[x][y] - , a[x + ][y] - a[x][y] + ) ; break ;
case 'E' : printf ("%d step(s) before a loop of %d step(s)\n" , a[x][y] - , a[x][y - ] - a[x][y] + ) ; break ;
case 'S' : printf ("%d step(s) before a loop of %d step(s)\n" , a[x][y] - , a[x - ][y] - a[x][y] + ) ; break ;
case 'W' : printf ("%d step(s) before a loop of %d step(s)\n" , a[x][y] - , a[x][y + ] - a[x][y] + ) ; break ;
}
}
void solve ()
{
x = , y = o ;
bool flag = ;
char temp ;
a[x][y] = ;
while (x >= && x <= row && y >= && y <= col) {
switch (map[x][y])
{
case 'N' : x-- ; if (a[x][y]) flag = ;
if (!flag)
a[x][y] = a[x + ][y] + ;
else
temp = 'N' ; break ;
case 'E' : y++ ; if (a[x][y]) flag = ;
if (!flag)
a[x][y] = a[x][y - ] + ;
else
temp = 'E' ; break ;
case 'S' : x++ ; if (a[x][y]) flag = ;
if (!flag)
a[x][y] = a[x - ][y] + ;
else
temp = 'S' ; break ;
case 'W' : y-- ; if (a[x][y]) flag = ;
if (!flag)
a[x][y] = a[x][y + ] + ;
else
temp = 'W' ; break ;
}
if (flag) {
loop (temp) ;
break ;
}
}
if (!flag)
printf ("%d step(s) to exit\n" , a[x][y] - ) ;
}
int main ()
{
// freopen ("a.txt" , "r" , stdin) ;
while (~ scanf ("%d%d%d" , &row , &col , &o)) {
if (row == && col == && o == )
break ;
memset (a , , sizeof(a)) ;
for (int i = ; i <= row ; i++)
for (int j = ; j <= col ; j++)
cin >> map[i][j] ; solve () ;
/* for (int i = 1 ; i <= row ; i++) {
for (int j = 1 ; j <= col ; j++) {
printf ("%d " , a[i][j]) ;
}
puts ("") ;
}
printf ("\n\n") ;*/
}
return ;
}

Robot Motion(imitate)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. 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 ...

  3. POJ 1573 Robot Motion(BFS)

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

  4. 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 ...

  5. POJ 1573 Robot Motion

    Robot Motion Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12978   Accepted: 6290 Des ...

  6. 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 ...

  7. POJ1573——Robot Motion

    Robot Motion Description A robot has been programmed to follow the instructions in its path. Instruc ...

  8. hdoj 1035 Robot Motion

    Robot Motion Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Tota ...

  9. HDU-1035 Robot Motion

    http://acm.hdu.edu.cn/showproblem.php?pid=1035 Robot Motion Time Limit: 2000/1000 MS (Java/Others)   ...

随机推荐

  1. (转)shell实例手册

    原文地址:http://hi.baidu.com/quanzhou722/item/f4a4f3c9eb37f02d46d5c0d9 实在是太好的资料了,不得不转 shell实例手册 0说明{ 手册制 ...

  2. PRML读书会第四章 Linear Models for Classification(贝叶斯marginalization、Fisher线性判别、感知机、概率生成和判别模型、逻辑回归)

    主讲人 planktonli planktonli(1027753147) 19:52:28 现在我们就开始讲第四章,第四章的内容是关于 线性分类模型,主要内容有四点:1) Fisher准则的分类,以 ...

  3. WP&Win10开发: RichTextBlock实现富文本并处理换行

    思路:1.构建字典.2.在字符串中匹配字典的key,将匹配到的key转换成对应的value3.将替换后的字符串,转化成xaml形式,加载该xaml以实现富文本. 代码如下: private Parag ...

  4. php中的错误级别

    在php编程过程中,大家一定会遇到或多或少的错误提醒,也正是这些错误提示,指引我们编写更加干净的代码,今天先写出我们主要列出的错误类型,先挖坑,写关于php错误与异常的相关知识,慢慢填坑.    De ...

  5. centos6.5上安装Openfire 4.0.3

    更新时间:2016年11月9日 00:18:27 博主的安装环境 物理机:        Win7 SP1 64位 ip:192.168.111.1    (用于安装spark 2.8.1) VM虚拟 ...

  6. JS面向对象概述

    这部分内容还是比较难理解的,像借用构造函数这种方法,实际工作中还是很常见的,不过对于后面的寄生理解还有点困难,只能慢慢学习了. 思维导图

  7. 三维数组——与 宝玉QQ群讨论交流之二

    宝玉 12:27:35 这几天看了大部分大家交的作业,发现一个主要问题还是卡在对三维数组的理解上,之前把三维数组类比成三维空间可能会造成误导 宝玉 12:27:45 其实鞠老师解释的很好: 三维数组 ...

  8. Spring MVC框架

    这个Spring Web MVC 框架提供了模型视图控制器的架构,这种结构能够被用来开发灵活的和松耦合的Web应用程序. 这种MVC模式能够将应用程序分离成不同的层面,(输入逻辑,业务逻辑,UI逻辑) ...

  9. .NET Core 在Visual Studio 2015 下的使用-MSDN

    .NET Core RC2 现已推出,这是真正的"候选发布"而非 RC1 Beta 冒充的候选发布(如果是那样,请考虑发布后出现的所有更改).当前,围绕 .NET Core 的开发 ...

  10. Cocos2d-X3.0 刨根问底(八)----- 场景(Scene)、层(Layer)相关源码分析

    本章节我们重点分析Cocos2d-x3.0与 场景.层相关的源码.这部分源码集中在 libcocos2d –> layers_scenes_transitions_nodes目录下面 我先发个截 ...