传送门 思路 题目读完，很容易想到DFS的思路，很快就可以撸一个标准的dfs解法： 记录当前最大值maxn，每次遍历到最后一个job就更新maxn到ans中。通过函数体中第一个if来维护“最小的最大值”。思路很清晰，可惜会超时。 根据题设，k和n的范围都是[1,12]，这个dfs的复杂度是O(kn)的复杂度,也就是12……
来自分类：做题
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