zoukankan      html  css  js  c++  java
  • poj 1573 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)

    机器人走网格,模拟


     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 using namespace std;
     6 int main()
     7 {
     8     int a,b,c; //行数,列数,机器人所在的列
     9     char str[15][15];
    10     bool vis[15][15];
    11     int count[15][15];
    12     while(cin>>a>>b>>c,a||b||c)
    13     {
    14         int step=0;
    15         char ch;
    16         memset(str,'O',sizeof(str)); //初始化为字符‘O’
    17         memset(vis,false,sizeof(vis)); //标记是否访问过
    18         memset(count,0,sizeof(count)); //记录步数
    19         for(int i=1; i<=a; i++)
    20         {
    21             for(int j=1; j<=b; j++)
    22                 cin>>str[i][j];
    23         }
    24         int i=1,j=c; //机器人所在的位置
    25         while(1)
    26         {
    27             if(vis[i][j]==true) //这个点以走过,构成了循环
    28             {
    29                 cout<<count[i][j]<<" step(s) before a loop of "<<step-count[i][j]<<" step(s)"<<endl;
    30                 break;
    31             }
    32             ch=str[i][j];
    33             if(ch=='O') //已经走出
    34             {
    35                 cout<<step<<" step(s) to exit"<<endl;
    36                 break;
    37             }
    38             count[i][j]=step;
    39             vis[i][j]=true;
    40             switch(ch)
    41             {
    42             case 'N':
    43                 i--;
    44                 step++;
    45                 continue;
    46             case 'S':
    47                 i++;
    48                 step++;
    49                 continue;
    50             case 'E':
    51                 j++;
    52                 step++;
    53                 continue;
    54             case 'W':
    55                 j--;
    56                 step++;
    57                 continue;
    58             }
    59         }
    60     }
    61     return 0;
    62 }
    View Code


  • 相关阅读:
    求幂运算、多项式乘法及Horner法则的应用
    JAVA泛型中的类型擦除及为什么不支持泛型数组
    关于递归的理解及递归表达式复杂度分析(以求解最大公约数为例)
    随机序列生成算法---生成前N个整数的一组随机序列
    Windows 与 Linux下关于端口不能访问的问题
    Netty 实现HTTP文件服务器
    字符数组转换成数字
    字符串反转的进一步应用----单词反转
    递归算法编程整数因子分解问题的递归算法
    数据返回[数据库基础]——图解JOIN
  • 原文地址:https://www.cnblogs.com/cxbky/p/4954140.html
Copyright © 2011-2022 走看看