Too much thinking! Too much annoying.】的更多相关文章

Annoying problem Problem's Link: http://acm.hdu.edu.cn/showproblem.php?pid=5296 Mean: 给你一个有根树和一个节点集合S,初始S为空,有q个操作,每个操作要么从树中选一个结点加入到S中(不删除树中节点),要么从S集合中删除一个结点.你需要从树中选一些边组成集合E,E中的边能够是S中的节点连通.对于每一个操作,输出操作后E中所有边的边权之和. analyse: 首先是构图,把树看作一个无向图,使用邻接表存图. 处理出…
nothing could be more annoying!!!!! lost!failed!…
C. Annoying Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Alice got an array of length nn as a birthday present once again! This is the third year in a row! And what is more disa…
log里看不出问题,直接客户端就disconnected. gdb 挂了也不会停住,继续跑得跟正常人似的 再连根本不正常的了. 硬件: a , 主板CPU更换过 b,USB3.0 软件: 无有更换,但是我有三个版本 9.0  9.1  DEBUG 找虫子...The Annoying Bug.... 1,my first app(connected with pc) 2,auto update 3,reboot and power off. selenium…
Annoying problem 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5293 Description Coco has a tree, whose vertices are conveniently labeled by 1,2,-,n. There are m chain on the tree, Each chain has a certain weight. Coco would like to pick out some ch…
Annoying problem 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5296 Description Coco has a tree, whose nodes are conveniently labeled by 1,2,-,n, which has n-1 edge,each edge has a weight. An existing set S is initially empty. Now there are two kin…
Annoying problem Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 203    Accepted Submission(s): 60 Problem Description Coco has a tree, whose nodes are conveniently labeled by 1,2,-,n, which h…
C. Annoying Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Alice got an array of length nn as a birthday present once again! This is the third year in a row! And what is more disa…
Annoying problem Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 483    Accepted Submission(s): 148 Problem Description Coco has a tree, whose nodes are conveniently labeled by 1,2,-,n, which…
题解链接 Annoying problem Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 480    Accepted Submission(s): 146 Problem Description Coco has a tree, whose nodes are conveniently labeled by 1,2,-,n, w…
Annoying painting tool 题目描述 Maybe you wonder what an annoying painting tool is? First of all, the painting tool we speak of supports only black and white. Therefore, a picture consists of a rectangular area of pixels, which are either black or white.…
Annoying problem Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1006    Accepted Submission(s): 330 Problem Description Coco has a tree, whose nodes are conveniently labeled by 1,2,…,n, which…
Annoying problem Problem Description Coco has a tree, whose nodes are conveniently labeled by 1,2,…,n, which has n-1 edge,each edge has a weight. An existing set S is initially empty.Now there are two kinds of operation: 1 x: If the node x is not in…
dfs一遍得到每一个节点的dfs序,对于要插入的节点x分两种情况考虑: 1,假设x能够在集合中的某些点之间,找到左边和右边距离x近期的两个点,即DFS序小于x的DFS序最大点,和大于x的DFS序最小的点...... 2.假设x在集合中的点某一側,则找距离x的dfs序最小和最大的点 将x插入这个集合最少要走的距离为 dist[x]-dist[LCA(left,x)]-dist[LCA(right,x)]+dist[LCA(left,right)] 删除同理 Annoying problem Tim…
题意: 给一棵n个节点的树,再给q个操作,初始集合S为空,每个操作要在一个集合S中删除或增加某些点,输出每次操作后:要使得集合中任意两点互可达所耗最小需要多少权值.(记住只能利用原来给的树边.给的树边已经有向.10万个点,10万个操作) 思路:只能用 O(nlogn)的复杂度.官方题解: 重点也就是要找到集合S中的以x和y为端点一条链,使得操作点u到达这条链是最近的.删除也是这样,找到这条链,删除u到这链的路长. 步骤: (1)记录从根遍历的DFS序. (2)计算每个点到根的路径所经过边的权之和…
LCA+RMQ.挺不错的一道题目. 思路是如何通过LCA维护费用.当加入新的点u是,费用增量为dis[u]-dis[lca(u, lower_u)] - dis[lca(u, greater_u)] + dis[lca(lower_u, greater_u)].若beg[u]大于当前最大值或小于最小值,lower_u=min of current, greater_u = max of current. /* 5296 */ #include <iostream> #include <s…
Bayesian inference Using Gibbs Sampling 允许用户指定复杂的多层模型,并可使用MCMC算法来估计模型中的未知参数. We use DAGs to specify models. 这里只涉及简单的贝叶斯网络,具体学习可见: Carnegie Mellon University course 10-708, Spring 2017, Probabilistic Graphical Models Ref: http://www.cnblogs.com/Dzhouq…
I am now in great demand for an opportunity to yearn for, the ability to express myself, in a maximum scale to let other people know about me, and my feeling as well. First: a good atmosphere could affect someone in a good way, and otherwise may be c…
Eclipse -> Preferences -> General -> Startup and Shutdown. -Uncheck RSE UI. Eclipse -> Preferences -> Remote Systems. -Uncheck Re-open Remote Systems view to previous state. Update your Eclipse to 4.3.1 (at least) due to a bug on previous v…
//#pragma comment(linker, "/STACK:1677721600") #include <map> #include <set> #include <stack> #include <queue> #include <cmath> #include <ctime> #include <vector> #include <cstdio> #include <cct…
题目链接:http://codeforces.com/contest/1009/problem/C 解题心得: 题意就是一个初始全为0长度为n的数列,m此操作,每次给你两个数x.d,你需要在数列中选一个位置pos,然后对于每一个位置i的数num[i]-x+d*dis(i,pos).最后要求这个数列平均值最大. 其实不需要记录这个数列中的每个数是多少,只需要记录总和就行了.如果d为正数pos选在数列越中间的位置最后的总和越大.所以需要分类讨论d的正负和数列长度的奇偶,然后用等差通项求和公式就行了.…
[链接] 我是链接,点我呀:) [题意] 题意 [题解] 其实就是让你最后这n个数字的和最大. 加上的x没有关系.因为肯定都是加上n个x 所以直接加上就可以了 主要在于如何选取j 显然我们要找到一个位置j. 然后pre[j]+aft[j]的值最大(pre[j]=1+2+3+...+j-1,aft类似 (这是在d大于0的时候,小于0的时候找pre[j]+aft[j]最小的就好) [代码] #include <bits/stdc++.h> #define ll long long using na…
http://codeforces.com/group/1EzrFFyOc0/contest/1009/problem/C 题意:原本有一个n个0的数组a[],你对它进行m次操作,每次操作让a[j]+=x+d*(dish(i,j))(dish(i,j)代表abs(i-j)).其中i是任意的.让你求经过这m次操作所能得到数组的平均值最大为多少.其实这里有一点贪心的思想,就是我们要尽量的让a[j]最大,那么和x是没有关系的,初始时数组和为ans=0;那么以后每次操作ans肯定会+=n*x;重点是对d…
题意 一个长度为n的数组(初始全为0),进行m次操作. 操作:给你x,d,你任意挑选一个 i (1~n),每个数字加上 x+|i-j|*d( j 表示对应数字的下标) 问m次操作后的最大算术平均值为多少? 题解 首先对于x,每次数组的和sum都增加n*x.(Σ|i-j|)*d跟我们选的 i 有关系,如果d>0,我们就让 Σ|i-j| 尽量大,如果d<0,我们就让 Σ|i-j| 尽量小.Σ|i-j|的最大值就是 i 取0或n的时候,Σ|i-j|的最小值就是 i 取n/2的时候.维护sum就行了…
第二篇 前言 本篇是和GIF相关的一个UIImage的分类.主要提供了三个方法: + (UIImage *)sd_animatedGIFNamed:(NSString *)name ----- 根据名称获取图片 + (UIImage *)sd_animatedGIFWithData:(NSData *)data ----- 根据NSData获取图片 - (UIImage *)sd_animatedImageByScalingAndCroppingToSize:(CGSize)size -----…
在微软的官方网站上有关于如何在SharePoint当中使用JS创建一个简单的文本文件的例子,经过我的思考我觉得结合Html5特性的浏览器,是完全可以通过JS来读取到文件的内容的(这一部分的内容请大家自行去了解),进而可以实现一个纯的JS的文件上传的例子,经过我的不断的实践,终于可以实现纯JS的文件上传了,当然还是参考了如下的文章. 限制有2个: 1.文件不能太大1.5MB以内完全是没有问题,这是JSOM的限制,没有办法突破. 2.浏览器必须是完全支持HTML5的,Chrome完全可以,IE必须1…
0x00. 烧盘 使用UltraISO(破解版)烧录到U盘里,设置电脑从U盘启动,即可安装. 安装时可能出现not COM32R image的命令行,“boot:” 后面直接输入live即可解决问题. 0x01. 顺序结构安装 这步不用多讲,就按顺序安装就行. 0x02. 更新 sudo apt-get update sudo apt-get upgrade 系统安装完需要更新,这两行命令提供系统更新.有人估计会说要更换源,公司网速可以的话,也没必要. 0x03. aptitude sudo a…
本文转自:https://www.exceptionnotfound.net/writing-custom-middleware-in-asp-net-core-1-0/ One of the new features from ASP.NET Core 1.0 is the idea of Middleware. Middleware are components of an application that examine the requests responses coming in t…
I found a related TechNet Blog that shed some light on the subject:http://blogs.technet.com/b/tune_in_to_windows_intune/archive/2013/11/13/replace-certificates-on-adfs-3-0.aspx According to this document, after setting the Service Communications Cert…
最近在学深度学习相关的东西,在网上搜集到了一些不错的资料,现在汇总一下: Free Online Books  by Yoshua Bengio, Ian Goodfellow and Aaron Courville Neural Networks and Deep Learning42 by Michael Nielsen Deep Learning27 by Microsoft Research Deep Learning Tutorial23 by LISA lab, University…