每日一题（三十）：DFS

题目描述：
The doggie found a bone in an ancient maze, which fascinated him a lot. However, when he picked it up, the maze began to shake, and the doggie could feel the ground sinking. He realized that the bone was a trap, and he tried desperately to get out of this maze.
The maze was a rectangle with sizes N by M. There was a door in the maze. At the beginning, the door was closed and it would open at the T-th second for a short period of time (less than 1 second). Therefore the doggie had to arrive at the door on exactly the T-th second. In every second, he could move one block to one of the upper, lower, left and right neighboring blocks. Once he entered a block, the ground of this block would start to sink and disappear in the next second. He could not stay at one block for more than one second, nor could he move into a visited block. Can the poor doggie survive? Please help him.

输入：
The input consists of multiple test cases. The first line of each test case contains three integers N, M, and T (1 < N, M < 7; 0 < T < 50), which denote the sizes of the maze and the time at which the door will open, respectively. The next N lines give the maze layout, with each line containing M characters. A character is one of the following:
'X': a block of wall, which the doggie cannot enter; 
'S': the start point of the doggie; 
'D': the Door; or
'.': an empty block.
The input is terminated with three 0's. This test case is not to be processed. 

输出：
For each test case, print in one line "YES" if the doggie can survive, or "NO" otherwise.

样例输入：
4 4 5
S.X.
..X.
..XD
....
3 4 5
S.X.
..X.
...D
0 0 0
样例输出：
NO
YES
 

#include<stdio.h>
char maze[8][8];
int n, m, t;
bool success;
int go[][2] = {
	1,0,
	-1,0,
	0,1,
	0,-1
};
void DFS(int x, int y, int time) { //递归形式的深度优先搜索
	for (int i = 0; i < 4; i++) {
		int nx = x + go[i][0];
		int ny = y + go[i][1];
		if (nx<1 || nx>n || ny<1 || ny>m)
			continue;
		if (maze[nx][ny] == 'X')
			continue;  //若该位置为墙，则跳过
		if (maze[nx][ny] == 'D') {
			if (time + 1 == t) {
				success = true;
				return;
			}
			else continue;  //否则该状态的后续状态不可能为答案（经过的点不能再经过）
		}
		maze[nx][ny] = 'X';
		DFS(nx, ny, time + 1);//递归扩展该状态，所用时间增加
		maze[nx][ny] = '.';//其后序状态全部遍历完毕，退回上层状态
		if(success)
			return;
	}
}
int main() {
	while (scanf("%d%d%d", &n, &m, &t) != EOF) {
		if (n == 0 && m == 0 && t == 0)
			break;
		for (int i = 1; i <= n; i++) {
			scanf("%s", maze[i] + 1);
		}
		success = false;
		int sx, sy;
		for (int i = 1; i <= n; i++) {    //寻找D的位置
			for (int j = 1; j <= m; j++) {
				if (maze[i][j] == 'D') {
					sx = i; 
					sy = j;
				}
			}
		}
		for (int i = 1; i <= n; i++) {    //寻找初始状态
			for (int j = 1; j <= m; j++) {
				if (maze[i][j] == 'S' && (i + j) % 2 == ((sx + sy) % 2 + t % 2) % 2) {
					//判断S与D的奇偶关系
					maze[i][j] = 'X';  //将起点标记为墙
					DFS(i, j, 0);      //递归扩展初始状态
				}
			}
		}
		puts(success == true ? "YES" : "NO");
	}
	return 0;
}