LCS
所属分类:其他
开发工具:C++
文件大小:1KB
下载次数:3
上传日期:2009-01-20 12:13:12
上 传 者:
lynn2009
说明: 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.
(Instead of finding the longest commonsubsequence, let us try to determine thelength of the LCS .)
文件列表:
LCS\LCS.cpp (2308, 2009-01-19)
LCS (0, 2009-01-20)
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