区间dp,枚举走完第一个子树之后回到根节点的位置. /*by SilverN*/ #include<algorithm> #include<iostream> #include<cstring> #include<cstdio> #include<cmath> #include<vector> using namespace std; const int mod=1e9; ; int read(){ ,f=;char ch=getch…

Exploring Pyramids Archaeologists have discovered a new set of hidden caves in one of the Egyptian pyramids. The decryption of ancient hieroglyphs on the walls nearby showed that the caves structure is as follows. There are n <tex2html_verbatim_mark>…

Exploring Pyramids 题目大意:给定一个欧拉序列(即每经过一个点,把这个点加入序列),问有多少种对应的多叉树 序列与树构造对应问题,考虑区间DP dp[i][j]表示序列i...j对应二叉树个数 初始i == j,dp[i][j] = 1 dp[i][j] = 0,i!=j 转移:dp[i][j] = sum(dp[i + 1][k - 1] * dp[k][j]),s[i] == s[j]  即考虑从i出发,在k这个位置回来,然后再从k出发,到j的时候回来 被MOD卡了一下 #…

Archaeologists have discovered a new set of hidden caves in one of the Egyptian pyramids. The decryption of ancient hieroglyphs on the walls nearby showed that the caves structure is as follows. There are n caves in a pyramid, connected by narrow pas…

http://codeforces.com/gym/101334 题意: 给出一棵多叉树,每个结点的任意两个子节点都有左右之分.从根结点开始,每次尽量往左走,走不通了就回溯,把遇到的字母顺次记录下来,可以得到一个序列. 思路:d[i][j]表示i~j的序列所对应的子树. 边界条件就是d[i][i]=1. 每次可以分为两个分支:d[i+1][k-1]和d[k][j]. 递归求解. #include<iostream> #include<algorithm> #include<c…

传送门 题目大意 看样例,懂题意 分析 实际就是个区间dp,我开始居然不会...详见代码(代码用的记忆化搜索) 代码 #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #include<cctype> #include<cmath> #include<cstdlib> #inclu…

题意:给定一个先序遍历序列,问符合条件的树的种类数 解题关键:枚举分割点进行dp,若符合条件一定为回文序列,可分治做,采用记忆化搜索的方法. 转移方程:$dp[i][j] = \sum {dp[i + 1][k - 1]*dp[k][j]} $ 令$dp[i][j]$表示i到j里数量 1.记忆化搜索 #include <bits/stdc++.h> using namespace std; typedef long long ll; const int mod=1e9; ll dp[][];…

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1517 久违的树形dp dp[l][r] 表示在l到r之间字符串形成的子树有多少种 然后要枚举最左树枝所到位置 假设是 i 那么从l+1到i-1 递归就是最左树枝的种类 然后乘上剩下的部分 剩下的部分i到r相当是去掉了最左树枝的又一个子树,递归就可以 代码: #include…

题意 给定一个DFS序列,问能有多少树与之对应. 思路 设输入序列为S,dp(i, j)为子序列Si, Si+1, --, Sj对应的树的个数,则边界条件为d(i, i) = 1,且Si != Sj时d(i, j) = 0(因为起点和终点应该是同一个点).在其他情况下,设第一个分支在Sk时回到树根(必须有Si == Sk),则这个分支对应的序列是Si+1, Si+2, -- , Sk-1,方案数为dp(i+1, k-1):其他分支访问序列为Sk, Sk+1, --, Sj,方案数为dp(k, j…

设d(i, j)为连续子序列[i, j]构成数的个数,因为遍历从根节点出发最终要回溯到根节点,所以边界情况是:d(i, i) = 1; 如果s[i] != s[j], d(i, j) = 0 假设第一个分支在Sk回到根节点,方案数为d(i+1, k-1) 其他分支访问从Sk到Sj,方案数为d(k, j) 根据乘法原理,d(i, j) = sum{d(i+1, k-1), d(k, j), i+2≤k≤j 且 Si = Sk = Sj} #include <bits/stdc++.h> usin…

题意:给定一个序列 问有多少棵树与之对应 题目连接:https://cn.vjudge.net/problem/UVALive-3516 对于这一序列  分两种2情况 当前分支 和 其它分支  用dfs 在当前层的dfs 只讨论当前分支  其他分支 dfs到下一层 构成分支的条件 即为str[i] == str[j] 即首和末相同 然后再去判断着一串上的其它字母 是否符合回溯的条件 #include <iostream> #include <cstdio> #include <…

f(i,j)=sum(f(i+1,k-1)*f(k,j) | i+2<=k<=j,Si=Sk=Sj). f(i+1,k-1)是划分出第一颗子树,f(k,j)是划分出剩下的子树. #include<cstdio> #include<cstring> using namespace std; typedef long long ll; #define MOD 1000000000ll char s[310]; ll f[310][310]; int n; ll dp(int…

-----------------------------最优化问题------------------------------------- ----------------------常规动态规划  SOJ1162 I-Keyboard  SOJ1685 Chopsticks SOJ1679 Gangsters SOJ2096 Maximum Submatrix  SOJ2111 littleken bg SOJ2142 Cow Exhibition  SOJ2505 The County…

仅供参考,还未运行程序,理解部分有误,请参考英文原版. 绿色部分非文章内容,是个人理解. 转载请注明:http://blog.csdn.net/raby_gyl/article/details/17471617 Chapter 4:Exploring Structure from  Motion Using OpenCV 在这一章,我们将讨论来至运动结构(Structure from Motion,SfM)的概念,或者从一个运动的相机拍摄到的图像中更好的推测提取出来的几何结构,使用OpenCV的…

Angular 1.5 introduced the .component() helper method, which is much simpler than the.directive() definition and advocates best practices and common default behaviours. Using .component() will allow developers to write in an Angular 2 style as well,…

Windows with C++ Exploring Fonts with DirectWrite and Modern C++ Kenny Kerr DirectWrite is an incredibly powerful text layout API. It powers practically all of the leading Windows applications and technologies, from the Windows Runtime (WinRT) implem…

Exploring Python Code Objects https://late.am/post/2012/03/26/exploring-python-code-objects.html Inspired by David Beazley's Keynote at PyCon, I've been digging around in code objects in Python lately. I don't have a particular axe to grind, nor some…

Exploring the 7 Different Types of Data Stories What makes a story truly data-driven? For one, the numbers aren’t caged in a sidebar graph. Instead, the data helps drive the narrative. Data can help narrate as many types of stories as there are angle…

Lesson 6: Exploring the World of Typefaces 这课提到的字体都是 英文 的. Article 1: More Google Web Fonts That Don't Suck 列出了很多 Google Fonts 上的好看的字体. FontFriend 这个工具可以改变网站的字体,让你看看各种字体在你的网站上到底是什么效果.很牛逼! Article 2: MyFonts Bestsellers These are the top 50 selling fo…

Lab 6 Exploring the Bash Shell Sequence 1: Directory and file organization 1. Log in as user student with the password student. 2. [student@stationX ~]$ pwd/home/student 3. [student@stationX ~]$ touch {report,memo,graph}_{sep,oct,nov,dec}_{a,b,c}_{1,…

Lab 10 Exploring Virtualization Goal: To explore the Xen virtualization environment and the creation of a Domain-U virtual machine. Sequence 1: Installing the Xen Virtualization Environment Deliverable: A Red Hat Enterprise Linux system running the X…

<a href='http://www.engineeringvillage.com/blog/document.url?mid=cpx_10ed754f14b5b7381b6M764b1017816338&database=cpx'>Exploring the use of a 3D virtual environment in chinese cultural transmission</a> <table border="0" cellspa…

项目地址:http://phototour.cs.washington.edu/ Photo Tourism是华盛顿大学的SFM重建的过程 Paper:Photo Tourism: Exploring Photo Collections in 3D 文章提出的方法如下: (1)提取SIFT特征点: (2)每两幅图像之间进行SIFT特征点进行匹配: 匹配过程中首先采用了KD-TREE的方法对最近邻的特征点进行匹配,然后采用了多视角几何进行限制.值得注意的是多视角的几何限制,在多视角的图像匹配中经常…

Exploring Mesh APIs using light switch example The light switch example is meant to showcase the APIs for the provisioner and provisionee roles and how a simple Bluetooth Mesh network may be set up and configured. The network consists of one static p…

Description Recently in Farland, a country in Asia, a famous scientist Mr. Log Archeo has discovered ancient pyramids. But unlike those in Egypt and Central America, they have triangular (not rectangular) foundation. That is, they are tetrahedrons in…

转自  http://bits-please.blogspot.com/2015/08   (需要FQ, 狗日的墙) In this blog post, we'll be exploring Qualcomm's TrustZone implementation, as present on Snapdragon SoCs. If you haven't already, you might want to read theprevious blog post, in which I go i…

自动类型安全的.NET标准REST库refit   在SCOTT HANSELMAN 博客上看到一个好东西<Exploring refit, an automatic type-safe REST library for .NET Standard>,他推荐了一个.NET标准1.4 的自动类型安全的REST库refit. refit 类似于Java的Retrofit,是一套RESTful架构的.NET客户端实现,基于特性,提供把REST API返回的数据转化为(Plain Ordinary C…

作者:Omar Alonso 会议:CIKM 2009 摘要: 截至目前(2009),通过提取文档中内嵌的时间信息来展现和聚类,这方面的工作并不多. 在这篇文章中,我们将提出一个“小插件”增添到现有的传统信息检索系统应用中,这个小附件能够探索多种时间信息. 使用时间实体提取技术,我们将展现时间表达如何被显示的生成以及如何被用于构建多粒度的时间线上.我们将讨论搜索结果的列表如何根据时间来聚类,并作为锚文本嵌入到时间线上,并且基于时间的文本聚类如何被利用来探索包含时间片段的搜索结果.同时,我们将描绘…

opengl insight Exploring Mobile vs. DesktopOpenGL Performance Jon McCaffrey 前面那些内容以前看过 应该写在谋篇帖子里了  Fullscreen effect pp优化的核心在于 减pass 因为这全是system mem sync read modify write... 1.全屏特效 如果不需要 neighboring pixelx 用frame buffer ftech 和base做到一起 带宽可以做到最低 2.pa…

  CS224N Assignment 1: Exploring Word Vectors (25 Points)¶ Welcome to CS224n! Before you start, make sure you read the README.txt in the same directory as this notebook. In [7]: # All Import Statements Defined Here # Note: Do not add to this list. #…