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
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)
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
char s[][];
int i,j,m,n,a,p[][];
while(scanf("%d %d %d%*c",&m,&n,&a)&&m&&n&&a)
{
memset(p,,sizeof(p));
for(i=; i<m; i++)
{
for(j=; j<n; j++)
scanf("%c",&s[i][j]);
getchar();
}
j=a-;
i=;
p[i][j]=;
while()
{
if(s[i][j]=='N')
{
if(i-<)
{
printf("%d step(s) to exit\n",p[i][j]);
break;
}
else if(p[i-][j])
{
printf("%d step(s) before a loop of %d step(s)\n",p[i-][j]-,p[i][j]+-p[i-][j]);
break;
}
else
{
p[i-][j]=p[i][j]+;
i--;
continue;
}
}
else if(s[i][j]=='S')
{
if(i+==m)
{
printf("%d step(s) to exit\n",p[i][j]);
break;
}
else if(p[i+][j])
{
printf("%d step(s) before a loop of %d step(s)\n",p[i+][j]-,p[i][j]+-p[i+][j]);
break;
}
else
{
p[i+][j]=p[i][j]+;
i++;
continue;
}
}
else if(s[i][j]=='W')
{
if(j-<)
{
printf("%d step(s) to exit\n",p[i][j]);
break;
}
else if(p[i][j-])
{
printf("%d step(s) before a loop of %d step(s)\n",p[i][j-]-,p[i][j]+-p[i][j-]);
break;
}
else
{
p[i][j-]=p[i][j]+;
j--;
continue;
}
}
else if(s[i][j]=='E')
{
if(j+==n)
{
printf("%d step(s) to exit\n",p[i][j]);
break;
}
else if(p[i][j+])
{
printf("%d step(s) before a loop of %d step(s)\n",p[i][j+]-,p[i][j]+-p[i][j+]);
break;
}
else
{
p[i][j+]=p[i][j]+;
j++;
continue;
}
}
}
}
return ;
}
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