[ACM] hdu 1035 Robot Motion (模拟或DFS）

Posted 2020-09-20 clnchanpin

Robot Motion

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)

 

Source

Mid-Central USA 1999

解题思路：

每一个地图位置上都有一个方向。到达当前坐标。依据当前坐标的方向进行下一步的行走，给出起始位置，模拟机器人行走。可能会逃脱地图。输出步数，也可能会陷入死循环，也就是某个坐标位置第二次遇到，输出陷入循环前走的步数以及循环里面走的步数。模拟一下。推断是否有循环，到达坐标x,y的步数用step[x][y]存储，假设下一步是合法的行走（即没有越界，也没有反复訪问），那么step[nextx][nexty]=step[x][y]+1,假设有循环，那么用单独一个变量duo来保存第二次到达某坐标位置所须要的步数，break掉。

代码：

#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
const int maxn=110;
char map[maxn][maxn];
int step[maxn][maxn];//到达当前点走的步数
bool visit[maxn][maxn];//是否已经訪问
int nextx,nexty;//下一个坐标位置
int n,m,p;//地图行列，開始位置的列数
bool loop;//推断走的路径是否出现环
int duo;//假设出现环。记录第二次走到该位置所须要的步数

bool escape(int x,int y)//推断是否逃出地图
{
    if(x<1||x>n||y<1||y>m)
        return true;
    return false;
}

void input(int n,int m)
{
    memset(step,0,sizeof(step));
    memset(visit,0,sizeof(visit));
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
        cin>>map[i][j];
}

void walk(int x,int y)//模拟行走
{
    visit[x][y]=1;
    while(1)
    {
        if(map[x][y]==‘E‘)//推断当前坐标方向，来确定下一个坐标位置
        {
            nextx=x;
            nexty=y+1;
        }
        else if(map[x][y]==‘W‘)
        {
            nextx=x;
            nexty=y-1;
        }
        else if(map[x][y]==‘N‘)
        {
            nextx=x-1;
            nexty=y;
        }
        else
        {
            nextx=x+1;
            nexty=y;
        }
        if(escape(nextx,nexty))//已经逃脱，则步数为step[x][y]+1,nextx,nexty为全局变量，不管能否逃脱，最后输出步数用nextx,nexty做參数比較方便
        {
            nextx=x;
            nexty=y;
            break;
        }
        else if(visit[nextx][nexty])//第二次訪问该位置
        {
            loop=1;//出现环
            duo=step[x][y]+1;
            break;
        }
        else//既没有逃脱地图。下一个位置也没有被訪问
        {
            visit[nextx][nexty]=1;
            step[nextx][nexty]=step[x][y]+1;//关键，下一个位置步数比前个位置步数多1
            x=nextx;//这里是为了连接while循环，注意看while循环里面的第一条if语句
            y=nexty;
        }
    }
}

int main()
{
    while(cin>>n>>m>>p&&(n||m||p))
    {
        input(n,m);
        loop=0;
        walk(1,p);
        if(!loop)//没有环
        {
            cout<<step[nextx][nexty]+1<<" step(s) to exit"<<endl;
        }
        else
            cout<<step[nextx][nexty]<<" step(s) before a loop of "<<duo-step[nextx][nexty]<<" step(s)"<<endl;
    }
    return 0;
}