题目代号:POJ 1573

题目链接:http://poj.org/problem?id=1573

Language:
Default
Robot Motion
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 14195   Accepted: 6827

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

题目大意:WESN分别代表四个方向,第三个数据代表第一行的第几个位置开始,如果能走出去则输出走了几步,如果不能走出去进入了循环则输出第几步进入了循环,循环有几步。按题目标准格式输出。

解题思路:水题,作个标记代表第几步,如果下一步被标记过了则退出循环输出,如果走出去了也退出循环。

AC 代码:

# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL; char a[][];
int b[][]; int main()
{
//freopen("in.txt", "r", stdin);
int n,m,k;
while(cin>>n>>m,n&&m)
{
cin>>k;
mem(a,);
mem(b,);
for(int i=;i<=n;i++)
cin>>a[i]+;
int x=,y=k;
b[x][y]=;
int ans=;
int flag=;
while()
{
if(a[x][y]=='W')y--;
else if(a[x][y]=='S')x++;
else if(a[x][y]=='E')y++;
else if(a[x][y]=='N')x--;
if(x==||x==n+||y==||y==m+)
{
printf("%d step(s) to exit\n",ans);
break;
}
else if(b[x][y])
{
printf("%d step(s) before a loop of %d step(s)\n",b[x][y]-,ans-b[x][y]+);
break;
}
else
{
b[x][y]=++ans;
}
}
}
return ;
}

POJ 1573 Robot Motion(模拟)的更多相关文章

  1. POJ 1573 Robot Motion 模拟 难度:0

    #define ONLINE_JUDGE #include<cstdio> #include <cstring> #include <algorithm> usin ...

  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【模拟题 写个while循环一直到机器人跳出来】

                                                                                                         ...

  4. POJ 1573 Robot Motion(BFS)

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

  5. POJ 1573 Robot Motion

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

  6. poj 1573 Robot Motion_模拟

    又是被自己的方向搞混了 题意:走出去和遇到之前走过的就输出. #include <cstdlib> #include <iostream> #include<cstdio ...

  7. 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 ...

  8. [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. ...

  9. PKU 1573 Robot Motion(简单模拟)

    原题大意:原题链接 给出一个矩阵(矩阵中的元素均为方向英文字母),和人的初始位置,问是否能根据这些英文字母走出矩阵.(因为有可能形成环而走不出去) 此题虽然属于水题,但是完全独立完成而且直接1A还是很 ...

随机推荐

  1. CM金丝雀Canary报错

    参考: https://www.cnblogs.com/barneywill/p/10400788.html CM金丝雀Canary报错 1 HDFS 金丝雀Canary 测试无法为 /tmp/.cl ...

  2. pycharm中ctrl + C复制, ctrl+A全选等快捷键失效

    原因是:在安装pycharm的时候也同时安装了vim插件,需要在settings  - > vim Emulation里将相关的handler改成 IDE

  3. 刷机,twrp,安装xposed

    首先明白几个名词: recovery模式,类似于pc端的PE系统,每个手机都有自带的rec,但不好用,最好自己刷一个,现在市面最好用的是twrp fastboot模式,比recovery更底层,进入f ...

  4. reload() 方法用于重新加载当前文档。配合Ajax异步请求。

    1. reload() 方法, reload() 方法用于重新加载当前文档.配合Ajax异步请求. http://www.w3school.com.cn/jsref/met_loc_reload.as ...

  5. 记一次神奇的codeforces

    今天有一场codeforces的div3,时间挺合适,于是就想打.结果发现rating超过1600就不能报名.虽然shzr好久不打CF了而且很菜,但是毕竟还是到了1600的,于是和ZUTTER_一起用 ...

  6. 【原创】Themida 2260 虚拟机 FISH 初探(一)

    标 题: [原创]Themida 2260 虚拟机 FISH 初探(一)作 者: xiaohang时 间: 2016-03-03,00:39:37链 接: http://bbs.pediy.com/s ...

  7. 84. Largest Rectangle in Histogram (JAVA)

    Given n non-negative integers representing the histogram's bar height where the width of each bar is ...

  8. Java RMI 最简单实例

    IHello.java import java.rmi.Remote; import java.rmi.RemoteException; public interface IHello extends ...

  9. oracle获取年、月、日

    --获取年 select extract(year from date'2011-05-17') year from dual; --获取月 select extract(month from dat ...

  10. Ubuntu分区方案(菜鸟方案、常用方案和进阶方案)

    菜鸟方案 “/”与swap两个分区就可以应付绝大多数的应用 常用方案 分为3个区 1. 挂载点/:主分区:安装系统和软件:大小为30G:分区格式为ext4: 2. 挂载点/home:逻辑分区:相当于“ ...