Problem Description
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.
Input
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.
Output
For each test case, print in one line “YES” if the doggie can survive, or “NO” otherwise.
Sample Input
4 4 5
S.X.
…X.
…XD
…
3 4 5
S.X.
…X.
…D
0 0 0
Sample Output
NO
YES
题意嘛，就是UZI（不知道的去问问度娘）要在规定时间结束的那一刻到达门口，也就是D点，而看时间的话，直接DFS可能会超时，那怎么办呢？如何才能最理想呢？
就这个题来想的话，可能有以下几点，
- 时间可以优化。想一下，假如走到某一点的时候，所用的时间已经超过了给定的时间，或者是剩余的步数也就是剩余的时间不够你到达门口，那还需要再搜索吗？
- 墙壁，这个看起来没有用，其实不然，如果墙壁数太多路太少，时间又多余可走的路程，那是不是也就没必要去判断了呢？
- 最后一点，状态！是否到达D点的标记，如果你明明已经到了终点，却没有将程序结束，是不是很坑？对的，设置一个标记去记录是否到达D点，到达且所用时间刚好等于题目所给的时间，那就return
（我做这个题的时候，想了将近2天吧（刚接触搜索）哈哈哈哈哈哈，感觉自己很沙雕）
