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LCS
- 这是一个合并两个文件的工程,用来将两个文件合并,相同部分保留一次,不同部分一次写入,其中利用了求最大公共子序列的算法-This is a merger of the two files works, used to merge two files, the same part of the reservation time, different parts of the write-once, which demand the greatest advantage of the common s
main3
- 使用c++,依据动态规划思路,解决在一个数组中,找到和最大的子序列-Using c++, based on dynamic programming ideas to solve in an array to find and the largest sub-sequence of
ex2subsequence
- 求取最大连续子序列之和与最大连续子序列,输入在input2.dat文件中,输出在result2文件中-Derive the maximum sum consecutive subsequence with the maximum continuous sub-sequence, enter the input2.dat file, the output file in the result2
maxlist
- 输入序列文件 输出第一行为最大连续子序列之和 第二行为最大连续子序列,注意这里可能有多个最大连续子序列,每一个最大连续子序列占一行-Input sequence file output of the first acts of the largest and second consecutive sub-sequences of the greatest acts of continuous sub-sequences, note that there may be more th
Suanfa_2
- java实现,寻找最大连续子序列完整算法,可以找出所有最大连续子序列-java implementation, a complete search algorithm for maximum consecutive subsequence, can find all of the largest sequence of consecutive sub-
lcs
- 大二课程算法设计技巧与分析的实验,实现最长公共子序列-Two course design skills and analysis algorithm, the longest public son experiment
the-max-sum
- 一串数字序列比如1,3,4,5,7,6……得到其中的最带不连续子序列(要求序列是严格递增的如1,3,4,5,7可以,5,7,6就不可以) 本代码即,最大不连续子序列问题,属于经典的动态规划问题适合初学者-A string of number sequences such as 1,3,4,5,7,6 ... ... be one of the best sequence with discontinuous (which requires a strictly increasing sequ
most-common-subsequence-string
- 算法设计类源码。使用动态规划的方式计算两个字符串的最大公共子序列。-Algorithm design class source code. Calculated using dynamic programming the two most common subsequence string.
shuangduandui
- 双端队列应用于优化算法,解决竞赛题:由题目所说,Mr Pote有n个袋子装豆子。Mr Pote从中进行挑选包装,要求找出连续一列袋子,使得它们的豆子总和s在满足s p<=k条件下最大 ,以保证被狗狗吃掉的最少,并且Mr .Pote装的豆子最多。最后输出狗狗吃掉的豆子数目,即s p的结果。 抽象后表述为:给定n,p,k,以及n项数列[wi],要求找出其一个连续子序列,使得其和s在满足s p<=k的条件下最大,输出s整除p的结果。-Double-ended queue used in
MaxSum-(2)
- 用动态规划算法实现最大m子段和问题,给定N个整数组成的序列,以及一个正整数m,要求确定序列的m个不相交子段,使这m个子段的总和达到最大-Dynamic programming algorithm with the maximum m sub-section and the problem, given a sequence of N integers, and a positive integer m, required to determine the sequence of m disjoi
Maximum-Increasing-subsequences
- 最大递增子序列,动态规划经典算法 设L=<a1,a2,…,an>是n个不同的实数的序列,L的递增子序列是这样一个子序列Lin=<aK1,ak2,…,akm>,其中k1<k2<…<km且aK1<ak2<…<akm。求最大的m值。-Maximum increment sequence, the classic dynamic programming algorithm set L = <a1,a2,…,an> Are n d
maxsubsequence
- 输入一组数字序列后能够找到它的一组上升的最大长度的子序列。输出长度并输出此序列-Input a set of digital sequence can find it after a group of rising maximum length son sequence. Output length and output the sequence
sub-sequence
- 首先开设一个空间比较大的数组,从数组尾部开始。比较最后一个数和最 初设置在数组中的一个最大值,使得最小子序列及最小子序列初始化。记录在当前位置的最小子序列数,接着,继续与在其前面相邻的数进行比较。如果后者大于前者,者比较二者携带记录的最小子序列数的大小,如果后者持有的最小子序列数大于前者,则以后者为基准,最小子序列保持不变。反之,则将后者的子序列数自加一,将其改为前所持有的最小子序列数。继续重复前面的动作,最后得出最长不下降子序列的数目及内容。 -First to open a larg
BZZXL
- 不增子序列,求一序列的最大不增子序列长度-No increase in promoter sequences, find a sequence does not increase sub-sequence length
zixulie
- 求两个字符串的最长公共子序列X 的一个子序列是相应于X 下标序列{1, 2, …, m}的一个子序列,求解两个序列的所有 子序列中长度最大的,例如输入:pear, peach 输出:pea。 -Seeking two strings of the longest common subsequence
nyoj16
- 经典算法,最大单调递增子序列,查看最多能嵌入多少个矩形-The classical algorithm monotonically incrementing subsequence view can be embedded in the number of rectangular
MaxSum
- 求子序列最大值的实现算法-Subsequence maximum! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !
LCS
- 用Java实现的LCS最大公共子序列算法,输入从文件中读取ASCII码-LCS common subsequence algorithm implemented in Java, the input is read from the file in ASCII
LCS
- 求解所有的最大公共子序列,用于比较两个各字符串-All of LCS
how-to-get-ALL-LCS-document
- 求所有最大公共子序列的算法实现,小数据可以实现,大数据目前好像不行-For all of the biggest public subsequence algorithm implementation, small data can be realized, big data currently seems to be no