poj1573 Robot Motion

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题目链接：

http://poj.org/problem?id=1573

题目：

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

Output

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)

Source

代码为：

#include<cstdio>
#include<cstring>
const int maxn=10+10;
char map[maxn][maxn];
bool vis[maxn][maxn];
int step[maxn][maxn];
int row,colume,start;
int ex,ey;
int check(int x,int y)
{
    if(x>=1&&x<=row&&y>=1&&y<=colume)
        return 1;
    return 0;
}
int main()
{
    char str[maxn];
    int currentx,currenty,pos,ans;
    ind:while(scanf("%d%d%d",&row,&colume,&start)!=EOF)
    {
        pos=0;
        ans=1;
        memset(step,0,sizeof(step));
        memset(vis,false,sizeof(vis));
        if(row==0&&colume==0&&start==0)  return 0;
        for(int i=1;i<=row;i++)
        {
            scanf("%s",str);
            for(int j=1;j<=colume;j++)
              map[i][j]=str[j-1];
        }
        currentx=1;
        currenty=start;
        while(!vis[currentx][currenty])
        {
            if(!check(currentx,currenty))
            {
                 printf("%d step(s) to exit\n",ans);
                 goto ind;
            }
            vis[currentx][currenty]=true;
            if(map[currentx][currenty]=='E')
            {
                if(currentx==1&¤ty==start)
                    step[1][currenty]=0;
                else
                    {
                        pos++;
                        ans++;
                        step[currentx][currenty]=pos;
                    }
                    currentx=currentx;
                    currenty=currenty+1;
            }
            else if(map[currentx][currenty]=='S')
            {

                if(currentx==1&¤ty==start)
                    step[1][currenty]=0;
                else
                   {pos++;
                   ans++;
                       step[currentx][currenty]=pos;
                   }

                    currentx=currentx+1;
                    currenty=currenty;
            }
            else if(map[currentx][currenty]=='W')
            {

                if(currentx==1&¤ty==start)
                    step[1][currenty]=0;
                else
                    {  pos++;
                        step[currentx][currenty]=pos;
                     ans++;
                    }
                    currentx=currentx;
                    currenty=currenty-1;
            }
            else
            {
                if(currentx==1&¤ty==start)
                    step[1][currenty]=0;
                else
                    {
                        pos++;
                        step[currentx][currenty]=pos;
                        ans++;
                    }
                    currentx=currentx-1;
                    currenty=currenty;

            }
        }
        printf("%d step(s) before a loop of %d step(s)\n",step[currentx][currenty],ans-step[currentx][currenty]);
        goto ind;
    }
    return 0;
}

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poj1573 Robot Motion

标签：des   style   class   blog   code   http   

原文地址：http://blog.csdn.net/u014303647/article/details/30226557