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 ; }