Part of the practice is many, but every thought out of one also more 5~10 points. The positive solution is not yet, the following are a variety of partial approaches:
Subtask 1:k=1
The length of the LCS is up to 1, which means that there is no j>i and a[j]>a[i] at the same time. is obviously an LDS, the tree-like array directly to ask.
Subtask 2:k=2
Up to two, which can be combined by two LCS, F[i][j] indicates that the first LCS ends with I, the second is the number of scenarios ending with J, and the transfer is obvious.
Subtask 3:k=2
Tree-like array optimized DP, complexity reduced from $o (n^3) $ to $o (n^2 \log N) $
Subtask 4,5:b<=8
DP Set dp:https://www.cnblogs.com/clnchanpin/p/7357564.html
Generally with "subsequence" appears at the same time, such as the longest rise from the sequence, the longest public self-sequence and so on.
Subtask 6, 7:
An obvious theorem: a sequence of LCS with a maximum of k means that the sequence can be composed of at least one K disjoint LDS.
[CTSC2017] longest ascending self-sequence (pseudo-puzzle) (Tree array +dp dp+ minimum cost maximum flow +johnson shortest path +yang_tableau)