填坑系列(p.248) 比较神 从两端枚举 最坏复杂度就成O(nlogn)了 #include<cstdio> #include<cstdlib> #include<algorithm> #include<iostream> #include<cstring> #include<string> using namespace std; void setIO(const string& a) { freopen((a + &qu…

题意: 如果一个序列的任意连续子序列中至少有一个只出现一次的元素  则称这个序列是 不无聊序列  输入一个n个元素的序列a   判断是不是不无聊序列 一开始想当然  以为只要 2位的子序列都满足即可  题目果然没有那么脑残   1212 就是一个反例.... 有一个重要的优化结论:   在总的序列找出一个  只出现一次的元素a[p]  如果没有 则无解   否则      继续递归寻找  a[L]-a[p-1]  和 a[p+1] -a[R]   !!! 再一次巧妙运用递归 其中 判断是否为 唯…

某个序列找到唯一元素后,判断被分成的两边的序列即可问题在于找到唯一元素连续序列,重复元素的问题:感觉很有一般性 查找相同元素用map,last,next存上一个相同元素的位置复杂度计算有点思考:记录last,next,那么对于一个元素,判断是否独立O(1)从头开始查找,最坏情况T(n)=T(n-1)+O(n)(O(n)是从头判断到尾) =O(n^2)两边开始T(n)=2T(n/2)+O(n)=T(nlogn) #include<cstdio> #include<map> using…

思路: 算法很显然: 一.在区间\([l,r]\)找到一个只出现一次的元素P(如果不存在,那么序列\(boring\)) 二.递归处理区间\([l,p-1]\)和区间\([p+1,r]\). 其关键在于如何找到一个只出现一次的元素P. 首先,我们得知道如何判断一个元素是不是只出现一次. 我们可以用\(STL\)中的\(map\)记录与当前元素值相同的上一个元素 or 下一个元素的位置,然后更新即可. 因为\(map\)的所有操作都是\(O(log_n)\)的,所以预处理的时间复杂度为\(O(nl…

4059: [Cerc2012]Non-boring sequences Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 440  Solved: 160[Submit][Status][Discuss] Description 我们害怕把这道题题面搞得太无聊了,所以我们决定让这题超短.一个序列被称为是不无聊的,仅当它的每个连续子序列存在一个独一无二的数字,即每个子序列里至少存在一个数字只出现一次.给定一个整数序列,请你判断它是不是不无聊的. In…

要快速在一段子序列中判断一个元素是否只出现一次 , 我们可以预处理出每个元素左边和右边最近的相同元素的位置 , 这样就可以 O( 1 ) 判断. 考虑一段序列 [ l , r ] , 假如我们找到了序列中唯一元素的位置 p , 那我们只需检查 [ l , p - 1 ] & [ p + 1 , r ] 是否 non-boring 即可 . 如何检查 序列 [ l , r ] 呢 ? 假如从左往右或者从右往左找 , 最坏情况下是 O( n ) , 总时间复杂度会变成 O( n² ) ; 假如我们从…

题链: https://vjudge.net/problem/UVA-1608#author=chenchonghan题解: 分治 如果一个区间[l,r]里面在p位置出现了一个只出现一次的元素,(如果不存在该元素,该区间就是boring的) 那么必然包含p的子区间都是non-boring的. 即如果存在boring的区间,必然是[l,p-1],[p+1,r]的子区间. 所以继续递归处理上面两个区间即可. (判断某个元素是否在区间里只出现一次,只需考虑它左边第一个与它相同的元素或它右边第一个相同的…

Problem UVA1608-Non-boring sequences Accept: 227  Submit: 2541Time Limit: 3000 mSec Problem Description We were afraid of making this problem statement too boring, so we decided to keep it short. A sequence is called non-boring if its every connected…

http://poj.org/problem?id=1776 题意: 有一个机器要完成N个作业, 给你一个N*N的矩阵, M[i][j]=1,表示完成第i个作业后不用重启机器,继续去完成第j个作业 M[i][j]=0,表示如果做完第i个作业,想要继续去做第j个作业,那么必须重启机器 对于任意两个作业都有M[i][j] = 1或者M[j][i] = 1. 求出完成这N个作业启动机器的最少次数,以及每次启动完成作业的数量和这些作业的顺序 初始时机器处于关闭状态. 将M当做图,就是找最少的路径条数覆盖…

Boring Class Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 900    Accepted Submission(s): 247 Problem Description Mr. Zstu and Mr. Hdu are taking a boring class , Mr. Zstu comes up with a prob…

Non-boring sequences 大致题意: 给你一个字符串,问你他的任一子串是否都包含一个唯一的字符 思路: 看似简单,实际一丁点思路都没有 后面看汝佳的讲解都看了好长时间 大概思路就是,先找一个唯一字符,然后递归这个字符左右两边的字符串,看他们是否满足,知道只含一个元素.但是万一那个唯一元素是最后一个元素呢???这样复杂度就是n方了,这个时候神奇的优化方法出现了!!!!从两边往中间找,最坏情况就是在中间!!!!! 因为Tn = 2*T(n/2) + O(n),最坏也就nlogn 这样…

Mr. Zstu and Mr. Hdu are taking a boring class , Mr. Zstu comes up with a problem to kill time, Mr. Hdu thinks it’s too easy, he solved it very quickly, what about you guys? Here is the problem: Give you two sequences L1,L2,...,Ln and R1,R2,...,Rn. Y…

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA. Write a function to find all the 10-letter-long seq…

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA. Write a function to find all the 10-letter-long seq…

Weilin Huang--[AAAI2016]Reading Scene Text in Deep Convolutional Sequences 目录 作者和相关链接 方法概括 创新点和贡献 方法细节 实验结果 问题讨论 总结与收获点 参考文献 作者和相关链接 论文下载 黄伟林主页 , 乔宇,汤晓欧 所有作者 方法概括 解决问题:单词识别 主要流程:maxout版的CNN提取特征,RNN(LSTM)进行分类,CTC对结果进行调整.整个流程端到端训练和测试,和白翔的CRNN(参考文献1)方法几…

hihoCoder #1430 : A Boring Problem(一琐繁题) 时间限制:1000ms 单点时限:1000ms 内存限制:256MB Description - 题目描述 As a student of the school of electronics engineering and computer science in Peking University, Kyle took the course named Advanced Algebra in his freshma…

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA. Write a function to find all the 10-letter-long seq…

Problem Introduction In this problem, your goal is to compute the length of a longest common subsequence of three sequences. Problem Description Task.Given three sequences \(A=(a_1,a_2,\cdots,a_n)\); \(B = (b_1, b_2,\cdots,b_m)\), and \(C=(c_1,c_2,\c…

序列是指有序的队列,重点在"有序". 一.Python中序列的分类 Python中的序列主要以下几种类型: 3种基本序列类型(Basic Sequence Types):list.tuple.range 专门处理文本的附加序列类型(Text Sequence Types):str 专门处理二进制数据的附加序列类型(Binary Sequence Types): bytes.bytearray.memoryview 按照序列是否可被改变分类: 可变序列: list 不可变序列:tuple…

cut -d " " -f 1 sequences.fa | tr -s "\n" "\t"| sed -s 's/>/\n/g' > sequences.tab while read id start end; do \ g=$(grep "$id" sequences.tab | cut -f 2 | cut -c $start-$end);\ echo ">$id";\ echo $…

此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授. PDF格式教材下载 Sequences and Series 本系列学习笔记PDF下载(Academia.edu) MOOCULUS-2 Solution Summary Suppose that $\left(a_n\right)$ is a sequence. To say that $\lim_{n\to \infty}…

5.3 Tuples and Sequences We saw that lists and strings have many common properties, e.g., indexing and slicing operations. They are two examples of sequence data types. Since Python is an evolving language, other sequence data types may be added. The…

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA. Write a function to find all the 10-letter-long seq…

Boring Counting Time Limit: 3000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述     In this problem you are given a number sequence P consisting of N integer and Pi is the ith element in the sequence. Now you task is to answer a list of queries, for each…

Boring counting Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others) Total Submission(s): 2811    Accepted Submission(s): 827 Problem Description In this problem we consider a rooted tree with N vertices. The vertices a…

http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=2610 Boring Counting Time Limit: 3000ms   Memory limit: 65536K  有疑问?点这里^_^ 题目描述     In this problem you are given a number sequence P consisting of N integer and Pi is the ith ele…

Boring counting Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others) Problem Description In this problem we consider a rooted tree with N vertices. The vertices are numbered from 1 to N, and vertex 1 represents the root…

All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA. Write a function to find all the 10-letter-long seq…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5324 题意:给你一个二维的序列,让你找出最长的第一维升第二维降的子序列(如果多个答案,输出字典序最小) 解:考虑从后往前dp,在u点你需要知道u点之后的比u的第一维小,第二维大的dp最大值 可以用分治枚举u点之后比u的第一维大的点,然后用树状数组查询比u的第二维小的点中dp最大的 具体是: dp[i]表示以 i 开头的最长子序列,从后往前更新. 更新u点时有u.dp=max(v.dp)+1:v满足v…

Boring counting Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?pid=3518 Mean: 给你一个字符串,求:至少出现了两次(无重叠)的子串的种类数. analyse: 后缀数组中height数组的运用,一般这个数组用得很少. 总体思路:分组统计的思想:将相同前缀的后缀分在一个组,然后对于1到len/2的每一个固定长度进行统计ans. 首先我们先求一遍后缀数组,并把height数组求出来.height数组代表的含…