模拟_poj1573_Robot Motion

http://poj.org/problem?id=1573

#include <iostream>
using namespace std;
/*
一个容易被忽略的test case
2 2 1
SW
EN

正确输出：
0 step(s) before a loop of 4 step(s)

一个数组存储命令Grid
另一个数组存储走过的步数，出界则直接输出num，遇到循环则输出num-path[r][c]
(如果走到了曾经走过的路径上，一定会进loop)
*/
char grid[15][15];
int path[15][15];
int main() {
	int R, C, c;
	while (cin >> R >> C >> c && (R+C+c)) {
		memset(grid, 0, sizeof(grid));
		for (int i = 1; i <= R; ++i)
			for (int j = 1; j <= C; ++j)
				cin >> grid[i][j];

		int r = 1;
		int num = 0;
		memset(path, -1, sizeof(path));//初始化

		bool flag = false;
		while (1 <= r && r <= R && 1 <= c && c <= C) {
            //是否走到了走过的路径上
			if (path[r][c] > -1) {
				flag = true;
				break;
			}
			else
				path[r][c] = num++;

			if (grid[r][c] == 'N')
				r--;
			else if (grid[r][c] == 'S')
				r++;
			else if (grid[r][c] == 'E')
				c++;
			else if (grid[r][c] == 'W')
				c--;
		}

		if (flag)
			cout << path[r][c] << " step(s) before a loop of "
			<< num - path[r][c] << " step(s)" << endl;
		else
			cout << num << " step(s) to exit" << endl;
	}
	return 0;
}