@[PumpKin36](/user/1012783)
显而易见的，可以使用。
by _qingshu_ @ 2024-03-04 15:21:25
可以是可以但是时间复杂度O($2^n^n$)即使加优化和剪枝也会T两个点~~除非……~~
by qusia_MC @ 2024-03-05 19:15:05
@[PumpKin36](/user/1012783)
by qusia_MC @ 2024-03-05 19:15:23
@[William2019](/user/787512) 谢谢啦
by PumpKin36 @ 2024-03-05 20:17:37
@[_qingshu_](/user/602803) 栓q
by PumpKin36 @ 2024-03-05 20:18:31
@[William2019](/user/787512) @[PumpKin36](/user/1012783)
显而易见的，因为所有的点都只能向左与向下走，且与前面状态无关，迭代加深搜索即可。时间复杂度 $\mathcal{O}(n+m)$。
by _qingshu_ @ 2024-03-05 20:22:24
哦，是向右
by _qingshu_ @ 2024-03-05 20:22:43
每次向下或右走每次递归2次一共nm次时间复杂度就是O(n^m)
by qusia_MC @ 2024-03-05 20:48:50
@[_qingshu_](/user/602803)
by qusia_MC @ 2024-03-05 20:49:09
@[William2019](/user/787512)
说了迭代加深搜索啊...每一个点走一次不就可以了？
复杂度上面写错了 $\mathcal{O}(n\times m)$
by _qingshu_ @ 2024-03-05 20:50:23