Robot Motion(imitate)
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 11065 | Accepted: 5378 |
Description
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
Output
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
- poj1573 Robot Motion
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12507 Accepted: 6070 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 ...
- POJ1573——Robot Motion
Robot Motion Description A robot has been programmed to follow the instructions in its path. Instruc ...
- hdoj 1035 Robot Motion
Robot Motion Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tota ...
- 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
- python实现简易数据库之一——存储和索引建立
最近没事做了一个数据库project,要求实现一个简单的数据库,能满足几个特定的查询,这里主要介绍一下我们的实现过程,代码放在过ithub,可参看这里.都说python的运行速度很慢,但因为时间比较急 ...
- 《android基于andFix的热修复方案》思路篇
1:需求背景 项目上线之后,发现BUG需要修复(比如安卓兼容性等测试难以发现的问题),频繁的更新影响用户体验 2:方案要求 静默下载,耗费流量少,打完补丁后立刻生效,不用重启apk 3:解决思路 3. ...
- grootJS ui控件定义
index13.html <html><head> <title>ui控件定义</title> <script src="jquery- ...
- 接受客户端传的inputstream类型转成string类型
public string GetReqContent() { var request = HttpContext.Current.Request; var contentStream = reque ...
- 【BZOJ1013】【JSOI2008】球形空间产生器sphere(高斯消元)
1013: [JSOI2008]球形空间产生器sphere Time Limit: 1 Sec Memory Limit: 162 MBSubmit: 1600 Solved: 860[Submi ...
- 16.C#初见Lambda表达式及表达式树(九章9.1-9.3)
在说明Lambda相关知识前,我们需要了解Lambda表达式常用于LINQ,那么我们来聊下LINQ. LINQ的基本功能就是创建操作管道,以及这些操作需要的任何状态.这些操作表示了各种关于数据的逻辑: ...
- Bootstrap3.0学习第十九轮(JavaScript插件——标签页)
详情请看 http://aehyok.com/Blog/Detail/46.html 个人网站地址:aehyok.com QQ 技术群号:206058845,验证码为:aehyok 本文文章链接:ht ...
- Sublime-jQueryDocs
Package Control Messages======================== jQueryDocs---------- This package shows a selected ...
- OC基础--分类(category) 和 协议(protocol)
OC 中的category分类文件相当于 C#中的部分类:OC 中的protocol协议文件(本质是头文件)相当于 C#中的接口.今天就简单说明一下OC中的这两个文件. 由于视频中的Xcode版本低, ...
- BZOJ-2326 数学作业 矩阵乘法快速幂+快速乘
2326: [HNOI2011]数学作业 Time Limit: 10 Sec Memory Limit: 128 MB Submit: 1564 Solved: 910 [Submit][Statu ...