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
Learn how to:
*
Tokenize a null-terminated string
*
Create a search and replace function for strings
*
Implement subtraction for string objects
* Use the vector, deque, and list sequence containers
*
Use the container adaptors stack, queue, and priority_queue
* Use the map, multimap, set, and multiset associative containers
*
Reverse, rotate, and shuffle a sequence
*
Create a function object
*
Use binders, negators, and iterator adapters
*
Read and write files
*
Use stream iterators to handle file I/O
*
Use exceptions to handle I/O errors
*
Create custom inserters and extractors
*
Format date, time, and numeric data
* Use facets and the localization library
*
Overload the [ ], ( ), and -> operators
*
Create an explicit constructor
*
And much, much more
標簽：
null-terminated
Tokenize
Create
string
上傳時間：
2014-01-18
上傳用戶：yph853211
