杭电 hdu 1035 Robot Motion

嘿嘿，一次AC……哦耶～～～

总得来说不算是难题，不过在处理坐标的时候出了问题，呃，这个代码中会有略微提到，因为比较简单，就不多说了，相信通过自己测试数据，也能理解的。

/* THE PROGRAM IS MADE BY PYY */
/*----------------------------------------------------------------------------//
	Copyright (c) 2011 panyanyany All rights reserved.

	URL   : http://acm.hdu.edu.cn/showproblem.php?pid=1035
	Name  : 1035 Robot Motion

	Date  : Wednesday, August 10, 2011
	Time Stage : 2 hours

	Result:
4381936	2011-08-10 21:22:25 Accepted 1035 0MS 212K 1666 B C++ pyy


Test Data:

Review:
//----------------------------------------------------------------------------*/

#include <stdio.h>
#include <string.h>

#define infinity	0x7f7f7f7f
#define minus_inf	0x8f8f8f8f

#define MAXSIZE 11

char c ;
int row, col, start, step, loop ;
char grip[MAXSIZE][MAXSIZE], used[MAXSIZE][MAXSIZE] ;
int route[MAXSIZE][MAXSIZE] ;

void dfs (int x, int y)
{
	++step ;
	if (used[x][y])
	{
		loop = step - route[x][y] ;
		step = route[x][y] - 1 ;
	}
	else if (1 <= x && x <= col && 1 <= y && y <= row)
	{
		used[x][y] = 1 ;
		route[x][y] = step ;
		if ((c = grip[x][y]) == 'E')
			dfs (x+1, y) ;
		else if (c == 'W')
			dfs (x-1, y) ;
		else if (c == 'S')
			dfs (x, y+1) ;
		else if (c == 'N')
			dfs (x, y-1) ;
	}
	else
	{
		--step ;
	}
}

int main ()
{
	int i, j ;
	while (scanf ("%d%d%d", &row, &col, &start), row | col | start)
	{
		getchar () ;
		for (i = 1 ; i <= row ; ++i)
		{
			for (j = 1 ; j <= col ; ++j)
				scanf ("%c", &grip[j][i]) ; // 此处注意j 和i 的顺序
			scanf ("%c", &c) ;
		}

		memset (used, 0, sizeof (used)) ;
		step = loop = 0 ;
		dfs (start, 1) ;

		printf ("%d step(s) ", step) ;
		if (loop)
			printf ("before a loop of %d step(s)\n", loop) ;
		else 
			printf ("to exit\n") ;
	}
	return 0 ;
}