1573 Robot Motion （简单题）

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#1573 #ACM模拟题 #poj

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本文介绍了一道关于机器人路径模拟的编程题目。该题目要求通过分析预设的指令网格，判断机器人是否能成功退出网格或者陷入循环，并给出了具体的实现代码。

Description

1573 Robot Motion - jolt2 - 正在成长的菜鸟

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.

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)

这是一道模拟题，只要按照题目的意思一步一步的实现，就能成功。

exit的条件是大于行、列，或者小于0，loop的条件是第一次出现走过的位置。

#include<stdio.h>
#include<string.h>
int main()
{
	char map[11][11],m[11][11];//map代表grid，m记录走过的地方（m可以不要，直接对map进行操作，但要记录位置比较麻烦）
	int row,col,f,i,flag,rr,cc;
	while(1)
	{		
		flag = 0;
		scanf("%d%d%d",&row, &col, &f);
		int r = 0,c = f - 1;
		if(0 == row && 0 == col)
			break;
		for(i = 0; i < row; i++)
			scanf("%s",map[i]);
		memset(m,0,sizeof(m));	
		i = 1;
		while(flag == 0)
		{	
			//rr、cc来记录对一个指令的行和列，flag标记是exit了还是循环了
			rr = r;
			cc = c;
			switch(map[r][c])
			{
			case 'E':				
				if( c + 1 >= col)
					flag = 1;
				else
				{
					m[r][c] = i++;
					if(m[r][++c] > 0)
						flag = 2;
				}
				break;
			case 'N':
				if( r - 1 < 0 )
					flag = 1;
				else
				{
					m[r][c] = i++;
					if(m[--r][c] > 0)
						flag = 2;
				}
				break;
			case 'W':
				if( c - 1 < 0)
					flag = 1;
				else
				{
					m[r][c] = i++;
					if(m[r][--c] > 0)
						flag = 2;
				}break;
			case 'S':
				if( r + 1 >= row)
					flag = 1;
				else
				{
					m[r][c] = i++;
					if(m[++r][c] > 0)
						flag = 2;
				}break;
			}
		}
		if(1 == flag)
			printf("%d step(s) to exit\n",i);
		else
			printf("%d step(s) before a loop of %d step(s)\n",m[r][c] - 1,m[rr][cc] - m[r][c] + 1 );

	}
	return 0;
}