Longest Increasing Subsequence(DP)】的更多相关文章

Longest Increasing Subsequence 描述 给出一组长度为n的序列,a1​,a2​,a3​,a4​...an​, 求出这个序列长度为k的严格递增子序列的个数 输入 第一行输入T组数据 T(0≤T≤10) 第二行输入序列大小n(1≤n≤100),长度k(1≤k≤n) 第三行输入n个数字ai​(0≤ai​≤1e9) 输出 数据规模很大, 答案请对1e9+7取模 输入样例 1  2 3 2 1 2 2 3 2 1 2 3 输出样例 1 2 3 思路 用dp[i][j]数组记录在…
public static int LIS(List<Integer> al) { int[] arr = new int[al.size()]; int lis = 0; arr[0] = 1; for (int i = 1; i < al.size(); i++) { if (al.get(i) > al.get(i - 1)) arr[i] = arr[i - 1] + 1; else arr[i] = 1; } for (int i : arr) { if (arr[i]…
链接:http://poj.org/problem?id=2533 题解 #include<iostream> using namespace std; ]; //存放数列 ]; //b[i]表示以a[i]为结尾的子序列的最大长度 int main(){ int n; scanf("%d",&n); ;i<n;i++) scanf("%d",&a[i]); dp[]=; ;i<n;i++){ ;j<i;j++) //对于…
Level:   Medium 题目描述: Given an unsorted array of integers, find the length of longest increasing subsequence. Example: Input: [10,9,2,5,3,7,101,18] Output: 4 Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4. N…
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 Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 38980   Accepted: 17119 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ...…
Language: Default Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 33986   Accepted: 14892 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric seq…
Longest Ordered Subsequence Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 34454   Accepted: 15135 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ...…
Greatest Common Increasing Subsequenc Problem Description This is a problem from ZOJ 2432.To make it easyer,you just need output the length of the subsequence.   Input Each sequence is described with M - its length (1 <= M <= 500) and M integer numb…
问题 说明该问题在生物学中的实际意义 Biological applications often need to compare the DNA of two (or more) different organisms. A strand of DNA consists of a string of molecules called bases, where the possible bases are adenine, guanine, cytosine, and thymine(腺嘌呤,鸟嘌…
Greatest Common Increasing Subsequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3460    Accepted Submission(s): 1092 Problem Description This is a problem from ZOJ 2432.To make it easyer,…
一.Description 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. For example, seq…
讲解摘自百度; 最长公共上升子序列(LCIS)的O(n^2)算法? 预备知识:动态规划的基本思想,LCS,LIS.? 问题:字符串a,字符串b,求a和b的LCIS(最长公共上升子序列).? 首先我们可以看到,这个问题具有相当多的重叠子问题.于是我们想到用DP搞.DP的首要任务是什么?定义状态.? 1定义状态F[i][j]表示以a串的前i个字符b串的前j个字符且以b[j]为结尾构成的LCIS的长度.? 为什么是这个而不是其他的状态定义?最重要的原因是我只会这个,还有一个原因是我知道这个定义能搞到平…
Time limit2000 ms Memory limit65536 kB 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 < ... < i…
LIS和LCS的结合. 容易写出方程,复杂度是nm2,但我们可以去掉一层没有必要的枚举,用一个变量val记录前一阶段的最优解,这样优化成nm. 1<=k<j,j增加1,k的上界也增加1,就可以考虑用变量优化去掉一层循环. 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 using namespace std; 5 const int maxn=505; 6 int n,m,T,an…
题意: 给一个数组,求严格递增的最长递增子序列的长度. 思路: 开销是一个额外的O(n)的数组.lower_bound(begin,end,val)的功能是:返回第一个大于等于val的地址. class Solution { public: int lengthOfLIS(vector<int>& nums) { ; int *p=new int[nums.size()]; p[]=nums[]; ; ; i<nums.size(); i++) { if(nums[i]>p…
Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. 思路:如果不用动态规划,在两个for循环的情况下,还得依次比较i,j间的每个字符,O(n3).使用动态规划,O(n2) char* longestPa…
题目链接 Problem Description 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…
问题 L: Common Subsequence 时间限制: 1 Sec  内存限制: 32 MB 提交: 70  解决: 40 [提交][状态][讨论版] 题目描述 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 = <…
http://poj.org/problem?id=2533 在经典不过的DP题目了.... #include <map> #include <set> #include <stack> #include <queue> #include <cmath> #include <ctime> #include <vector> #include <cstdio> #include <cctype> #i…
链接:http://acm.hdu.edu.cn/showproblem.php?pid=1159 题解 #include<iostream> #include<cstring> using namespace std; ],B[]; //存放字符串 ][]; //存放到字符串a第 i+1个字符,字符串b第 j+1个字符为止的最大长度公共子序列的长度 int main(){ while(~scanf("%s%s",A,B)){ int alen=strlen(A…
在 Dynamic Programming | Set 1 (Overlapping Subproblems Property) 和 Dynamic Programming | Set 2 (Optimal Substructure Property) 中我们已经讨论了重叠子问题和最优子结构性质,现在我们来看一个可以使用动态规划来解决的问题:最长上升子序列(Longest Increasing Subsequence(LIS)). 最长上升子序列问题,致力于在一个给定的序列中找到一个最长的子序列…
一.本文内容 最长递增子序列的两种动态规划算法实现,O(n^2)及O(nlogn).     二.问题描述 最长递增子序列:给定一个序列,从该序列找出最长的 升序/递增 子序列. 特点:1.子序列不要求连续: 2.子序列在原序列中按严格(strictly)升序排序: 3.最长递增子序列不唯一.   注:下文最长递增子序列用缩写LIS表示.   example: 0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15   对应的LIS: 0, 2,…
[LeetCode]673. Number of Longest Increasing Subsequence 解题报告(Python) 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/number-of-longest-increasing-subsequence/description/ 题目描述: Given an unsorted array of integers, find the number of longest inc…
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/longest-increasing-subsequence/description/ 题目描述 Given an unsorted array of integers, find the length of longest increasing subsequence. E…
Leetcode之动态规划(DP)专题-392. 判断子序列(Is Subsequence) 给定字符串 s 和 t ,判断 s 是否为 t 的子序列. 你可以认为 s 和 t 中仅包含英文小写字母.字符串 t 可能会很长(长度 ~= 500,000),而 s 是个短字符串(长度 <=100). 字符串的一个子序列是原始字符串删除一些(也可以不删除)字符而不改变剩余字符相对位置形成的新字符串.(例如,"ace"是"abcde"的一个子序列,而"aec…
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 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…
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…
引出: 问题描述:给出一个序列a1,a2,a3,a4,a5,a6,a7….an,求它的一个子序列(设为s1,s2,…sn),使得这个子序列满足这样的性质,s1<s2<s3<…<sn并且这个子序列的长度最长.输出这个最长的长度.(为了简化该类问题,我们将诸如最长下降子序列及最长不上升子序列等问题都看成同一个问题,其实仔细思考就会发现,这其实只是<符号定义上的问题,并不影响问题的实质)例如有一个序列:1  7  3  5  9  4  8,它的最长上升子序列就是 1 3 4 8…