LCS Algorithm, this is a c++ code for lcs(Longest Common Subsequence)
标签： Subsequence Algorithm Longest Common
上传时间： 2013-12-25
上传用户：李梦晗
用Java编写的maximum contiguous Subsequence sum algorithm 实现, 希望对大家有所帮助
标签： Subsequence contiguous algorithm maximum
上传时间： 2014-12-20
上传用户：kelimu
Longest Ordered Subsequence,acm必备习题
标签： Subsequence Longest Ordered acm
上传时间： 2014-10-27
上传用户：1159797854
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).
标签： Subsequence sequence Problem Longest
上传时间： 2016-12-08
上传用户：busterman
最长子序列,英文版,电子书 ACM文章1977+Algorithms+for+the+Longest+Common+Subsequence+Problem
标签： Subsequence Algorithms Longest Problem
上传时间： 2014-01-27
上传用户：zhuimenghuadie
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.
标签： the Subsequence determine Instead
上传时间： 2013-12-17
上传用户：evil
参考算法导论写的LCS算法，仿照STL的泛型风格，适用于多种STL容器中的各种类型数据构成的序列的最大公共子序列（Longest Common Subsequence）问题求解。
上传时间： 2014-11-22
上传用户：stvnash
