poj1573&&hdu1035 Robot Motion(模拟)
转载请注明出处:http://blog.csdn.net/u012860063?
viewmode=contents
题目链接:
HDU: pid=1035">http://acm.hdu.edu.cn/showproblem.php?pid=1035
POJ: id=1573">http://poj.org/problem? id=1573
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
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
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)
纯模拟题;
代码例如以下:
#include <iostream>
using namespace std;
#include <cstring>
#define M 1000
char mat[M][M];
int re[M][M];
int n, m, s;
int ans, i, j, k;
void F()
{
i = 1; j = s;
k = 0;
for(int l = 1; ; l++)
{
if(re[i][j] != 0)
{
ans = k - re[i][j];
break;
}
if(i>n || i<1 || j>m || j<1)
{
break;
}
if(mat[i][j] == 'N')
{
k++;
re[i][j] = k;
i-=1;
continue;
}
if(mat[i][j] == 'S')
{
k++;
re[i][j] = k;
i+=1; continue;
}
if(mat[i][j] == 'W')
{
k++;
re[i][j] = k;
j-=1;
continue;
}
if(mat[i][j] == 'E')
{
k++;
re[i][j] = k;
j+=1;
continue;
}
}
}
int main()
{
while(cin>>n>>m>>s)
{
ans = 0;
if(n == 0 && m == 0 && s == 0)
break;
memset(mat,0,sizeof(mat));
memset(re,0,sizeof(re));
for(i = 1; i <= n; i++)
{
for(j = 1; j <= m; j++)
{
cin>>mat[i][j];
}
}
F();
if(ans == 0)
cout<<k<<" step(s) to exit"<<endl;
else
cout<<re[i][j]-1<<" step(s) before a loop of "<<ans+1<<" step(s)"<<endl;
}
return 0;
}
poj1573&&hdu1035 Robot Motion(模拟)的更多相关文章
- HDU-1035 Robot Motion 模拟问题(水题)
题目链接:https://cn.vjudge.net/problem/HDU-1035 水题 代码 #include <cstdio> #include <map> int h ...
- [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
http://acm.hdu.edu.cn/showproblem.php?pid=1035 Robot Motion Time Limit: 2000/1000 MS (Java/Others) ...
- hdu1035 Robot Motion (DFS)
Robot Motion Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Tot ...
- 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' ...
- 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 ...
- HDU1035 Robot Motion
Problem Description A robot has been programmed to follow the instructions in its path. Instructions ...
- POJ 1573 Robot Motion 模拟 难度:0
#define ONLINE_JUDGE #include<cstdio> #include <cstring> #include <algorithm> usin ...
- poj1573 Robot Motion
Robot Motion Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 12507 Accepted: 6070 Des ...
随机推荐
- swift + xcode 新手上路
有用的参考博文: 视频教程: 如何创建第一个iPhone App - HelloWorldHelloWorld 熟悉xcode: http://www.cocoachina.com/swift/201 ...
- C#线程应用实例(part1) 之 BeginInvoke和EndInvoke
最近这个公司是做 winfrom 开发的 , 这段时间就好好的学学WCF , 公司框架什么的自己去琢磨! 这里主要写一些 winfrom 中 用到的一些陌生 技术 1.BeginInvoke 以前B ...
- 快捷查看dll的PublicKeyToken
@echo off d: cd D:\Win2003\Microsoft Visual Studio 10.0\VC\ call vcvarsall.bat x86 echo. if not '%1' ...
- js正则:零宽断言
JavaScript正则表达式零宽断言 var str="abnsdfZL1234nvcncZL123456kjlvjkl"var reg=/ZL(\d{4}|\d{6})(?!\ ...
- QT无标题窗口在任务栏显示关闭(增加系统菜单)
在对话框中使用了如下代码: setWindowFlags(Qt::FramelessWindowHint); 在任务栏上右键点击程序,不会弹出菜单,解决办法,使用下面代码: setWindowFlag ...
- c显示数字的LED(数字转LED)
实现这么一个函数:传入一个int值,在屏幕输出类似LED显示屏效果的字母拼图,例如: 输入1234567890,输出: 请注意每个字符的固定宽度和高度,两个数字间保留一个空格. 函数名:void LE ...
- Google瓦片地图算法解析
基本概念: 地图瓦片地址:http://mt2.google.cn/vt/lyrs=m@167000000&hl=zh-CN&gl=cn&x=420&y=193& ...
- IOS算法(三)之插入排序
直接插入排序(Insertion Sort)的基本思想是:每次将一个待排序的记录,按其keyword大小插入到前面已经排好序的子序列中的适当位置,直到所有记录插入完毕为止. 设数组为a[0-n-1]. ...
- 使用CAShapeLayer和UIBezierPath画一个自定义半圆弧button
通常我们使用系统自带的UIButton时,一般都是Rect矩形形式的,或则美工给出一张半圆弧的按钮,如图为一张半圆加三角形的按钮,而此时,如果给按钮添加点击事件时,响应事件依然为矩形区域,不符合我们的 ...
- android 中的 ViewPager+ Fragment
android的Viewpager 的各种经常的用法,朋友问我要过,所以就稍微总结一下, ViewPager + Fragment 经常用到 代码是从 actionbarsherlock 中提取 ...