标签：poj   c++   

Robot Motion

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.

3 6 5
NEESWE
WWWESS
SNWWWW
4 5 1
SESWE
EESNW
NWEEN
EWSEN
0 0 

10 step(s) to exit
3 step(s) before a loop of 8 step(s)

#include<iostream>
#include<cstring>
using namespace std;
char ch[105][105];
int ans;
int v[105][105],w[105][105];	
int a,b,c,t;

void dfs(int x,int y)
{
	if(t==1)  return ;
	v[x][y]=1;
	ans++;
	if(ch[x][y]=='N')  x-=1;
	else  if(ch[x][y]=='E')  y+=1;
	else  if(ch[x][y]=='S')  x+=1;
	else  if(ch[x][y]=='W')  y-=1;
	if(x<0||y<0||x>=a||y>=b)
	{
		t=1;
		printf("%d step(s) to exit\n",ans);
		return ;
	}
	else  if(v[x][y])
	{
		t=1;
		printf("%d step(s) before a loop of %d step(s)\n",w[x][y],ans-w[x][y]);
		return ;
	}
	w[x][y]=ans;
	if(!v[x][y])
		dfs(x,y);
}

int main()
{

	while(cin>>a>>b>>c)
	{
		if(a==0||b==0||c==0)   break;
		t=0;
		int i,j;
		for(i=0;i<a;i++)
			scanf("%s",ch[i]);

		memset(v,0,sizeof(v));
		ans=0;
		for(i=0;i<a;i++)
			for(j=0;j<b;j++)
			{
				if(j==c-1)
				{
					dfs(i,j);
				}
			}
	}

	return 0;
}

POJ 1573 Robot Motion 搜索

标签：poj   c++   

原文地址：http://blog.csdn.net/hyccfy/article/details/37967915