POJ1573,Robot Motion,模拟水题，繁

最新推荐文章于 2021-08-17 23:39:04 发布

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最新推荐文章于 2021-08-17 23:39:04 发布

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本文介绍了一个模拟机器人在网格中根据指令移动的问题。通过编程实现机器人移动并判断其是否能离开网格或者进入循环路径，同时计算相应的步骤数。

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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)

分析：

模拟水题，比较繁琐，代码还是一如既往地挫。。。

用一个字符数组grid存储每个格对应的指示，vis记录是否走过这个格，对于某个状态如果下一步要走的格子之前已经访问过，则记录下一格的横坐标纵坐标。

用int steps(int i_,int j_,int i,int j)函数算出从坐标为(i_,j_)的格子到(i,j)格要走的步数，这样调用这个函数就可以算出从起点到循环点要走的步数以及循环的步数。

要注意当起点即为循环点的情况时到达循环点前的步数为0。

给出这组数据。。。

> Try this test case:
> 2 2 1
> SW
> EN
> 
> The right answer should be:
> 0 step(s) before a loop of 4 step(s)

code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#define MAX 10
char grid[MAX+2][MAX+2];
int vis[MAX+2][MAX+2];
using namespace std;
int steps(int i_,int j_,int i,int j)
{
    int step=0,tmpi=i_,tmpj=j_;
    while(tmpi!=i||tmpj!=j||step==0)
    {
        if(grid[tmpi][tmpj]=='N')
        {
            tmpi--;
            step++;
        }
        else if(grid[tmpi][tmpj]=='W')
        {
            tmpj--;
            step++;
        }
        else if(grid[tmpi][tmpj]=='S')
        {
            tmpi++;
            step++;
        }
        else if(grid[tmpi][tmpj]=='E')
        {
            tmpj++;
            step++;
        }
    }
    return step;
}
int main()
{
    int r,c,ent,i,j,step,tmpi,tmpj;
    while(scanf("%d %d %d",&r,&c,&ent),r||c||ent)
    {
        step=0;
        memset(grid,'0',sizeof(grid));
        memset(vis,0,sizeof(vis));
        for(i=1;i<=r;i++)
        {
            getchar();
            for(j=1;j<=c;j++)
                scanf("%c",&grid[i][j]);
        }
        tmpi=1;
        tmpj=ent;
        while(1)
        {
            vis[tmpi][tmpj]=1;
            if(grid[tmpi][tmpj]=='N')
            {
                    step++;
                    if(grid[--tmpi][tmpj]=='0')
                    {
                        printf("%d step(s) to exit\n",step);
                        break;
                    }
                    else if(vis[tmpi][tmpj]==1)
                    {
                        if(tmpi==1&&tmpj==ent)
                            printf("%d step(s) before a loop of %d step(s)\n",0,steps(tmpi,tmpj,tmpi,tmpj));
                        else
                            printf("%d step(s) before a loop of %d step(s)\n",steps(1,ent,tmpi,tmpj),steps(tmpi,tmpj,tmpi,tmpj));
                        break;
                    }
            }
            else if(grid[tmpi][tmpj]=='W')
            {
                    step++;
                    if(grid[tmpi][--tmpj]=='0')
                    {
                        printf("%d step(s) to exit\n",step);
                        break;
                    }
                    else if(vis[tmpi][tmpj]==1)
                    {
                        if(tmpi==1&&tmpj==ent)
                            printf("%d step(s) before a loop of %d step(s)\n",0,steps(tmpi,tmpj,tmpi,tmpj));
                        else
                            printf("%d step(s) before a loop of %d step(s)\n",steps(1,ent,tmpi,tmpj),steps(tmpi,tmpj,tmpi,tmpj));
                        break;
                    }
            }
            else if(grid[tmpi][tmpj]=='S')
            {
                    step++;
                    if(grid[++tmpi][tmpj]=='0')
                    {
                        printf("%d step(s) to exit\n",step);
                        break;
                    }
                    else if(vis[tmpi][tmpj]==1)
                    {
                        if(tmpi==1&&tmpj==ent)
                            printf("%d step(s) before a loop of %d step(s)\n",0,steps(tmpi,tmpj,tmpi,tmpj));
                        else
                            printf("%d step(s) before a loop of %d step(s)\n",steps(1,ent,tmpi,tmpj),steps(tmpi,tmpj,tmpi,tmpj));
                        break;
                    }
            }
            else if(grid[tmpi][tmpj]=='E')
            {
                    step++;
                    if(grid[tmpi][++tmpj]=='0')
                    {
                        printf("%d step(s) to exit\n",step);
                        break;
                    }
                    else if(vis[tmpi][tmpj]==1)
                    {
                        if(tmpi==1&&tmpj==ent)
                            printf("%d step(s) before a loop of %d step(s)\n",0,steps(tmpi,tmpj,tmpi,tmpj));
                        else
                            printf("%d step(s) before a loop of %d step(s)\n",steps(1,ent,tmpi,tmpj),steps(tmpi,tmpj,tmpi,tmpj));
                        break;
                    }
            }
        }
    }
    return 0;
}