LIS LCS n^2和nlogn解法 以及LCIS
Ref: http://www.cnblogs.com/gj-Acit/p/3236384.html 首先介绍一下LIS和LCS的DP解法O(N^2) LCS:两个有序序列a和b,求他们公共子序列的最大长度 我们定义一个数组DP[i][j],表示的是a的前i项和b的前j项的最大公共子序列的长度...
codeforces 10 D. LCIS LCIS O(n^2)算法
题目链接给出两个序列, 求出他们的最长公共上升子序列。两层循环, 内层循环j, 外层i。 如果a[i] == b[j], 那么dp[j] = max(dp[j], dp[best]+1), best是一个指针, 指向小于j的元素中dp值最大并且b[best]的值小于a[i]的元素。如果a[i]>...
BestCoder Round #87 1003 LCIS[序列DP]
LCIS Accepts: 109 Submissions: 775 Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)问题描述Alex有两个序列a1a2...ana1,a2...
hdu_4718_The LCIS on the Tree(树链剖分+线段树合并)
题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=4718题意:给你一棵树,每个节点有一个值,然后任给树上的两点,问这两点的最长连续递增区间是多少题解:先树链剖分,然后结合线段树的区间合并来搞,注意的是要记录递增和递减两个状态,因为线段树的区间都是从根到子...
hdu3308LCIS(线段树,点更新,段查寻,查寻时一定要注意跨越时如何计算)
Problem Description Given n integers. You have two operations: U A B: replace the Ath number by B. (index counting from 0) Q A B: output the length o...
LCIS POJ 2172 Greatest Common Increasing Subsequence
题目传送门题意:LCIS(Longest Common Increasing Subsequence) 最长公共上升子序列分析:a[i] != b[j]: dp[i][j] = dp[i-1][j]; a[i]==b[j]: dp[j]=max(dp[j],dp[k]); (1<=k<...
HDU4512完美队形I && HDU1423 Greatest Common Increasing Subsequence (LCIS)
填坑的时候又到啦,校赛因为不会LCIS所以吃了大亏,这里要补起来。LCIS就是在两个串里找最长上升子序列,相关的博客有很多,这里自己就不写那么多了。http://www.cnblogs.com/jackge/archive/2013/05/16/3081793.htmlhttp://www.cnbl...
HDOJ 3308 LCIS (线段树)
题目:ProblemDescriptionGivennintegers.Youhavetwooperations:UAB:replacetheAthnumberbyB.(indexcountingfrom0)QAB:outputthelengthofthelongestconsecutiveincr...
hdu-3308 LCIS (线段树区间合并)
LCISTimeLimit:6000/2000MS(Java/Others) MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):8337 AcceptedSubmission(s):3566ProblemDescription...
Codeforces Beta Round #10 D. LCIS(DP&LCIS)
D.LCIStimelimitpertest1secondmemorylimitpertest256megabytesinputstandardinputoutputstandardoutputThisproblemdiffersfromonewhichwasontheonlinecontest.T...