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.
资源简介: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,...
上传时间: 2013-12-17
上传用户:evil
资源简介:最长子序列,英文版,电子书 ACM文章1977+Algorithms+for+the+Longest+Common+Subsequence+Problem
上传时间: 2014-01-27
上传用户:zhuimenghuadie
资源简介:LCS Algorithm, this is a c++ code for lcs(Longest Common Subsequence)
上传时间: 2013-12-25
上传用户:李梦晗
资源简介:Linphone is a web phone: it let you phone to your friends anywhere in the whole world, freely, simply by using the internet. The cost of the phone call is the cost that you spend connected to the internet.
上传时间: 2015-05-04
上传用户:牛布牛
资源简介:The information in this publication is believed to be accurate as of its publication date. Such information is subject to change without notice and The ATM Forum is not responsible for any errors. The ATM Forum does not assume any respons...
上传时间: 2015-09-16
上传用户:1109003457
资源简介:This the second tutorial of the Writing Device Drivers series. There seems to be a lot of interest in the topic, so this article will pick up where the first left off. The main focus of these articles will be to build up little by little th...
上传时间: 2016-01-28
上传用户:lmeeworm
资源简介:The CC1000 RF transceiver is very easy to interface with a microcontroller. The chip is configured using a three-wire bus, comprising of PCLK, PDATA and PALE signals.
上传时间: 2014-01-04
上传用户:c12228
资源简介:Document showing how to determine facial features and the exact face contour
上传时间: 2017-03-16
上传用户:wangdean1101
资源简介: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. ...
上传时间: 2016-12-08
上传用户:busterman
资源简介:ngrep strives to provide most of GNU grep s common features, applying them to the network layer. ngrep is a pcap-aware tool that will allow you to specify extended regular or hexadecimal expressions to match against data payloads of packets...
上传时间: 2014-01-15
上传用户:bcjtao