zoukankan      html  css  js  c++  java
  • hdu1035

    Robot Motion

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4452    Accepted Submission(s): 2057

    Problem 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
     
    Sample Output
    10 step(s) to exit 3 step(s) before a loop of 8 step(s)
     
     
     
     
    1. #include<stdio.h>
    2. #include<string.h>
    3. int step,n,m;
    4. int a[1010][1010];
    5. char map[11][11];
    6. void DFS(int x,int y)
    7. {
    8.     while(x>=0&&y>=0&&x<n&&y<m&&map[x][y]!='O')
    9.     {
    10.         if(map[x][y]=='N')
    11.         {
    12.             map[x][y]='O';
    13.             a[x][y]=++step;
    14.             x--;
    15.         }
    16.         else if(map[x][y]=='S')
    17.         {
    18.             map[x][y]='O';
    19.             a[x][y]=++step;
    20.             x++;
    21.         }
    22.         else if(map[x][y]=='E')
    23.         {
    24.             map[x][y]='O';
    25.             a[x][y]=++step;
    26.             y++;
    27.         }
    28.         else if(map[x][y]=='W')
    29.         {
    30.             map[x][y]='O';
    31.             a[x][y]=++step;
    32.             y--;
    33.         }
    34.     }
    35.         if(map[x][y]=='O')
    36.             printf("%d step(s) before a loop of %d step(s)\n",a[x][y]-1,step+1-a[x][y]);
    37.         else
    38.             printf("%d step(s) to exit\n",step);
    39. }
    40. int main()
    41. {
    42.     int i,k,x,y;
    43.     while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    44.     {
    45.         if(n==0||m==0)
    46.             break;
    47.         for(i=0;i<n;i++)
    48.             scanf("%s",map[i]);
    49.         x=0;
    50.         y=k-1;
    51.         step=0;
    52.         DFS(x,y);
    53.     }
    54.     return 0;
    55. }
     
  • 相关阅读:
    怎么判断自己在不在一家好公司?
    超全!互联网大厂的薪资和职级一览
    Nginx 又一牛 X 功能!流量拷贝
    时间管理之四象限法则
    罗永浩一个坑位卖60万脏钱背后:放下面子赚钱,才是成年人最大的体面
    2020 年 4月全国程序员工资出炉
    一次 SQL 查询优化原理分析(900W+ 数据,从 17s 到 300ms)
    “Hey Siri” 背后的黑科技大揭秘!
    一文讲透高薪的本质!
    python UnicodeDecodeError: 'gbk' codec can't decode byte 0x99 in position 87: illegal multibyte sequence异常解决
  • 原文地址:https://www.cnblogs.com/Deng1185246160/p/2950857.html
Copyright © 2011-2022 走看看