[hdu6991]Increasing Subsequence】的更多相关文章

令$f_{i}$​​表示以$i$​​为结尾的极长上升子序列个数,则有$f_{i}=\sum_{j<i,a_{j}<a_{i},\forall j<k<i,a_{k}\not\in [a_{j},a_{i}]}f_{j}$ (初始状态为前缀最小值处$f_{i}=1$,最终答案为后缀最大值处的$f_{i}$​之和) 暴力计算复杂度显然为$o(n^{2})$,无法通过 考虑分治计算,当递归到区间$[l,r]$时,需要求出仅考虑$[l,r]$内部的(包括转移的$j$)时的$f_{i}$ 具…
Given an unsorted array of integers, find the length of longest increasing subsequence. For example, Given [10, 9, 2, 5, 3, 7, 101, 18], The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than…
Longest Increasing Subsequence(LIS) 一个美丽的名字 非常经典的线性结构dp [朴素]:O(n^2) d(i)=max{0,d(j) :j<i&&a[j]<a[i]}+1 直接两个for [二分查找优化]:O(n^2) g(i):d值为i的最小的a  每次更新然后lower_bound即可 [大于等于] lower_boundReturn iterator to lower bound Returns an iterator pointing…
Given a sequence of integers, find the longest increasing subsequence (LIS). You code should return the length of the LIS. Have you met this question in a real interview?     Example For [5, 4, 1, 2, 3], the LIS  is [1, 2, 3], return 3 For [4, 2, 4,…
Given a sequence of integers, find the longest increasing subsequence (LIS). You code should return the length of the LIS. Example For [5, 4, 1, 2, 3], the LIS  is [1, 2, 3], return 3 For [4, 2, 4, 5, 3, 7], the LIS is [4, 4, 5, 7], return 4 Challeng…
Given an unsorted array of integers, find the length of longest increasing subsequence. For example,Given [10, 9, 2, 5, 3, 7, 101, 18],The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than o…
Longest Increasing Subsequence Given an unsorted array of integers, find the length of longest increasing subsequence. For example,Given [10, 9, 2, 5, 3, 7, 101, 18],The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Not…
传送门 The task is to find the length of the longest subsequence in a given array of integers such that all elements of the subsequence are sorted in ascending order. For example, the length of the LIS for { 15, 27, 14, 38, 26, 55, 46, 65, 85 } is 6 and…
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<j&&b[k]<b[j]) 打印路径时按照b[i]来输出 收获:理解不是很深入,推荐资料: 最长公共上升子序列(LCIS)的O(n^2)算法 最长公共上升子序列的另一个O(mn)的算法…
题目: Given an unsorted array of integers, find the length of longest increasing subsequence. For example,Given [10, 9, 2, 5, 3, 7, 101, 18],The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more th…