Problem B:Longest Ordered Subsequence
A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the Subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered Subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered Subsequences are of length 4, e. g., (1, 3, 5, 8).
標簽：
Subsequence
sequence
Problem
Longest
上傳時間：
2016-12-08
上傳用戶：busterman
Instead of finding the longest common
Subsequence, let us try to determine the
length of the LCS.
Then tracking back to find the LCS.
Consider a1a2…am and b1b2…bn.
Case 1: am=bn. The LCS must contain am,
we have to find the LCS of a1a2…am-1 and
b1b2…bn-1.
Case 2: am≠bn. Wehave to find the LCS of
a1a2…am-1 and b1b2…bn, and a1a2…am and
b b b
b1b2…bn-1
Let A = a1 a2 … am and B = b1 b2 … bn
Let Li j denote the length of the longest i,g g
common Subsequence of a1 a2 … ai and b1 b2
… bj.
Li,j = Li-1,j-1 + 1 if ai=bj
max{ L L } a≠b i-1,j, i,j-1 if ai≠j
L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
標簽：
the
Subsequence
determine
Instead
上傳時間：
2013-12-17
上傳用戶：evil
