zoukankan      html  css  js  c++  java
  • POJ 1573 && hdu 1035(简单模拟)

    POJ 1573 && hdu 1035(简单模拟)

    感谢kb神

    题目链接:http://poj.org/problem?id=1573

                 http://acm.hdu.edu.cn/showproblem.php?pid=1035

    Robot Motion
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 7893   Accepted: 3813

    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)
    #include<stdio.h>
    #include<string.h>
    
    char map[15][15];
    int dis[15][15];
    
    int main()
    {
        int n,m,s;
        int i,j;
        int si,sj;
        int step;
        int step1;
        while(scanf("%d%d%d%*c",&n,&m,&s)!=EOF && m)
        {
            step=0;step1=0;
            memset(dis,0,sizeof(dis));
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=m;j++)
                    scanf("%c",&map[i][j]);
                getchar();
            }
            si=1;sj=s;
            while(si>=1 && si<=n && sj>=1 && sj<=m &&!dis[si][sj])
            {
                step++;
                dis[si][sj]=1;
                if(map[si][sj]=='W')
                    sj-=1;
                else if(map[si][sj]=='E')
                    sj+=1;
                else if(map[si][sj]=='S')
                    si+=1;
                else if(map[si][sj]=='N')
                    si-=1;
            } 
            if(!(si>=1 && si<=n) || !(sj>=1 && sj<=m))
                printf("%d step(s) to exit\n",step);
            else
            {
                memset(dis,0,sizeof(dis));
                while(!dis[si][sj])
                {
                    step1++;
                    dis[si][sj]=1;
                    if(map[si][sj]=='W')
                        sj-=1;
                    else if(map[si][sj]=='E')
                        sj+=1;
                    else if(map[si][sj]=='S')
                        si+=1;
                    else if(map[si][sj]=='N')
                    si-=1;
                }
                step=step-step1;
                printf("%d step(s) before a loop of %d step(s)\n",step,step1);
            }        
        }
        return 0;
    }
  • 相关阅读:
    Jenkins安装以及邮件配置
    TUXEDO管理命令总结
    QTP的退出函数
    QTP 启动应用软件方法
    Robot framework 和RIDE 关系
    使用pycharm编写和运行RF脚本
    刷题-力扣-102. 二叉树的层序遍历
    刷题-力扣-反转链表
    刷题-牛客-判断链表中是否有环
    刷题-力扣-278. 第一个错误的版本
  • 原文地址:https://www.cnblogs.com/freezhan/p/2636443.html
Copyright © 2011-2022 走看看