HDU-Robot Motion-机器人移动

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#c++ #Robot Motion

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本文探讨了如何通过编程实现机器人在网格中的路径导航，并检测其是否能够成功离开网格或进入循环路径。详细介绍了输入数据格式，算法流程，以及输出结果的两种可能情况。

问题及代码：

Problem A Robot Motion

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 7 Accepted Submission(s) : 5

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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)

/*    
*Copyright (c)2015,烟台大学计算机与控制工程学院    
*All rights reserved.    
*文件名称：HDU.cpp    
*作    者：单昕昕    
*完成日期：2015年2月25日    
*版 本 号：v1.0        
*/ 
#include<iostream>
#include<string>
#include<cstdio>
using namespace std;
int main()
{
    string str[100];
    char flag;
    int row,line,place,loop,i,m,n,k,p,q;
    while(cin>>row>>line>>place&&(row!=0)&&(line!=0)&&(place!=0))
     {
        for(i=0; i<row; i++)
        {
            cin>>str[i];
        }
        flag=str[0][place-1];
        m=k=loop=0;
        n=place-1;
        while(0<=m&&m<row&&0<=n&&n<line)
        {
            p=m;
            q=n;
            flag=str[m][n];
            if(flag!='E'&&flag!='W'&&flag!='S'&&flag!='N')
            {
                loop=1;
                break;
            }
            str[m][n]=1+k;
            k++;
            switch(flag)
            {
            case 'N':
                m--;
                break;
            case 'S':
                m++;
                break;
            case 'E':
                n++;
                break;
            case 'W':
                n--;
                break;
            }
        }
        if(loop)
            cout<<str[p][q]-1<<" step(s) before a loop of "<<k+1-str[p][q]<<" step(s)"<<endl;
        else
            printf("%d step(s) to exit\n",str[p][q]);
    }
    return 0;
}

运行结果：

知识点总结：

根据题目给出的方向，判断机器人能否走出。

学习心得：

我不知道为什么最后一个输出语句如果相应改成cout语句就会出错，求大神指教~~~

下面上出问题的代码：

#include<iostream>
#include<string>
#include<cstdio>
using namespace std;
int main()
{
    string str[100];
    char flag;
    int row,line,place,loop,i,m,n,k,p,q;
    while(cin>>row>>line>>place&&(row!=0)&&(line!=0)&&(place!=0))
     {
        for(i=0; i<row; i++)
        {
            cin>>str[i];
        }
        flag=str[0][place-1];
        m=k=loop=0;
        n=place-1;
        while(0<=m&&m<row&&0<=n&&n<line)
        {
            p=m;
            q=n;
            flag=str[m][n];
            if(flag!='E'&&flag!='W'&&flag!='S'&&flag!='N')
            {
                loop=1;
                break;
            }
            str[m][n]=1+k;
            k++;
            switch(flag)
            {
            case 'N':
                m--;
                break;
            case 'S':
                m++;
                break;
            case 'E':
                n++;
                break;
            case 'W':
                n--;
                break;
            }
        }
        if(loop)
            cout<<str[p][q]-1<<" step(s) before a loop of "<<k+1-str[p][q]<<" step(s)"<<endl;
        else
            cout<<int (str[p][q])<<" step(s) to exit"<<endl;
    }
    return 0;
}

额，需要加一个强制类型转换。。