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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。-reference Introduction to write the algorithm LCS algorithm, modeled on the STL Generic style. STL is applicable to many types of containers data constitute the
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LCS Longest (maximum) common subsequence
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求X和Y两个字符串的最长公共子序列。（X和Y字符串分别存放在ASCII码文件2X.txt和2Y.txt中)
要求：根据动态规划法基本设计思想，将实现任务的C语言程序代码和运行结果填写在试卷上。
-Two strings X and Y find the longest common subsequence. (X and Y strings are stored in ASCII files 2X.txt and 2Y.txt in) requirements: According to
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计算两个字符串的最长公共子字符串的长度，使用动态规划。-Calculation of two of the longest common sub-string length of string, the use of dynamic programming.
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最长子序列,英文版,电子书 ACM文章1977+Algorithms+for+the+Longest+Common+Subsequence+Problem-The longest sequence, in English, e-books ACM article 1977+ Algorithms+ For+ The+ Longest+ Common+ Subsequence+ Problem
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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 h
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算法的一种，找出两个字符序列中最长的共同子序列，并输出。-An algorithm to identify the two character sequence of the longest common subsequence, and output.
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Longest Common Subsequence Problem. Solution both Iterative and Recursive.
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finds the longest common subsequence.
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序列Z=<B，C，D，B>是序列X=<A，B，C，B，D，A，B>的子序列，相应的递增下标序列为<2，3，5，7>。
一般地，给定一个序列X=<x1,x2,…,xm>，则另一个序列Z=<z1,z2,…,zk>是X的子序列，是指存在一个严格递增的下标序列〈i1,i2,…,ik〉使得对于所有j=1,2,…,k使Z中第j个元素zj与X中第ij个元素相同。
给定2个序列X和Y，当另一序列Z既是X的子序列又是Y的子序列时，称Z
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一个最长公共子序列的实现算法，编译完全正确-One of the longest common subsequence algorithm，compile completely correct
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Constrained LCS. Given two sequences X, Y and a constrained sequence P, a sequence Z is a constrained longest common subsequence for X and Y with respect to P if Z is the longest subsequence of X and Y such that P is a subsequence of Z.
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Longest common subsequence problem. It is dynamic approach based problem.
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算法分析的最长公共子序列问题 求出其中的最优解 可在c++环境中做-Algorithm analysis of the longest common subsequence problem find the optimal solution which can be done in c++ environment
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实现两个不同序列求出他们的最大公共子序列-To achieve common subsequence
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利用动态规划法求解次最长公共子序列问题。-The use of dynamic programming method for solving second longest common subsequence problem.
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用动态规划实现查找两个数列之间的最长公共子序列-Realization of dynamic programming to find the number of columns between the two longest common subsequence
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最长公共子序列问题 第一步 最优子结构: 第二步：建立递归式 第三步：计算最优值 -The longest common subsequence problem
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poj 1458
A subsequence of a given sequence is the given sequence with some elements (possible none) left out. Given a sequence X = < x1, x2, ..., xm > another sequence Z = < z1, z2, ..., zk > is a subsequence of X if there exists a stri
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VC++实现最长公共子序列的查找定位，简单易懂-The longest common subsequence
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