seg

题目链接： https://www.luogu.org/problem/P3313 这道题目就是树链剖分+线段树动态开点。 然后做这道题目之前我们先来看一道不考虑树链剖分之后完全相同的线段树动态开点的题目： https://www.cnblogs.com/codedecision/p/11791200.html 然后你就会发现这就是树链剖分+上题的线段树处理。 然后这道题目就变得很简单。 实现代码如下： #include <bits/stdc++.h> using namespace std; #define INF (1<<29) const int maxn = 100010; int fa[maxn], dep[maxn], size[maxn], son[maxn], top[maxn], seg[maxn], seg_cnt, rev[maxn]; vector<int> g[maxn]; void dfs1(int u, int p) { size[u] = 1; for (vector<int>::iterator it = g[u].begin(); it != g[u].end(); it ++) { int v = (*it); if (v == p) continue; fa[v] = u; dep[v] = dep[u] + 1; dfs1(v, u);

目录 一、关于树链剖分 二、树链剖分实现流程 二、树链剖分具体实现 1.需要表示的变量 2.储存一棵树 3.第一次遍历，处理fa，dep，size，son数组 4.第二次遍历，处理top，seg，rev数组 5.初始化线段树 6.单点修改 7.区间修改---以x为根结点的子树内节点的值都加val 8.区间修改---节点x到节点y的最短路径中同时加val 9.区间查询---以x为根结点的子树内节点的值的和 10.区间查询---节点x到节点y的最短路径中节点的和 11.区间查询---节点x到节点y的最短路径中的最大值/最小值 三、例题 例1： P3384 【模板】树链剖分 一、关于树链剖分 你的好盆友最近抛给你这样一个难题 (无中生友) ： " 一棵树由n个节点，每个节点都有一个权值w，现在想让你对这棵树完成下列操作： 1.把节点u的权值改为t 2.询问节点u到节点v的权值和 3.节点u到v的最大值 " 你看了看题目，发现这就是树链剖分的板子题... 好吧，那如果你不会树链剖分呢？ ... 于是你的朋友告诉你这是树链剖分，并因为你不会树链剖分把你嘲讽了 (开玩笑而已啦) ... 只观察这个问题的三个操作，你惊讶的发现这是线段树所擅长的事情，即单点修改，区间查询。 实际上，如果这棵树退化成一条链，那么你完全可以用线段树来解决这个问题。 你思考了一下，得出了树链剖分是什么东西： 树链剖分

题目链接： luogu 题目分析： 把每个人当成一个三元组 \([l_i, r_i, v_i]\) 考虑每个人对哪个能力区间 \([L, R]\) 有贡献 应该是左端点在 \([l_i, v_i]\) ，右端点在 \([v_i, r_i]\) 的区间 拍到一个二维平面上，求最多有多少个矩形交一起 线段树维护扫描线即可 代码： #include<bits/stdc++.h> #define N (600000 + 10) using namespace std; inline int read() { int cnt = 0, f = 1; char c = getchar(); while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();} while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + (c ^ 48); c = getchar();} return cnt * f; } int n, Y[N], cnt, ans = 0, ansx, ansy, tot; int l[N], r[N], v[N]; struct node2 { int x, Y1, Y2, tag; }seg[N << 1]; struct node { int l, r; int tag, gmax

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I keep getting the follow error in my Apache log: [Wed Sep 18 17:59:20 2013] [notice] Apache/2.2.22 (Ubuntu) PHP/5.3.10-1ubuntu3.8 with Suhosin-Patch configured -- resuming normal operations [Wed Sep 18 18:06:30 2013] [notice] child pid 7505 exit signal Segmentation fault (11), possible coredump in /etc/apache2 [Wed Sep 18 18:06:35 2013] [notice] child pid 7497 exit signal Segmentation fault (11), possible coredump in /etc/apache2 [Wed Sep 18 18:13:53 2013] [notice] child pid 7501 exit signal Segmentation fault (11), possible coredump in

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have the Linkedin company URL as follows, http://in.linkedin.com/company/abb and the company ID for ABB limited is 277579 . Essentially you can also reach ABB through http://www.linkedin.com/company/277579 . But if I have only http://in.linkedin.com/company/abb . Is it possible to get the company ID through this URL? Does parsing the URL help? Seeking any good method to get the company ID. Is there any other way to do without using Linkedin API? Hope I am clear with my question. Any help would be appreciated. 回答1: Type in your company name

题目链接： https://www.luogu.org/problem/P2146 本题涉及算法： 树链剖分； 线段树（区间更新及求和，涉及懒惰标记） 然后对于每次 install x ，需要将 x 到 1 的路径上面的点全都置为1。 那么在置为1之前统计一下节点数量 num1， 在置为1之后统计一下节点数量 num2， 答案就是 num2 - num1 （当然，也可以通过节点深度 dep[x] 来获得节点数量）。 对于每次 unistall x ，需要将 x 为根的子树上面的点全都置为0。 那么在置为0之前统计一下权值为1的节点数量 num1， 在置为0之后统计一下权值为1的节点数量 num2， 答案就是 num1-num2 （当然， num2 其实就等于 0 ）。 实现代码如下： #include <bits/stdc++.h> using namespace std; #define INF (1<<29) const int maxn = 100010; int fa[maxn], dep[maxn], size[maxn], son[maxn], top[maxn], seg[maxn], seg_cnt, rev[maxn], n, sumv[maxn<<2], lazy[maxn<<2]; vector<int> g[maxn]; void dfs1(int u,

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I know that this question has been asked before, but several new versions have been released since then. Is it possible to handle a dblclick event for creating a new appointment in the calendar without having to modify the fullcalendar.js file? It would be great to handle this extension in a separate file together with my other tweaks. Thanks in advance! /Adam 回答1: Adam's post at https://code.google.com/p/fullcalendar/issues/detail?id=231 uses this solution. whenever possible i like to keep things out of the core of fullcalendar, and have

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 选择语言 中文(简体) 日语 英语 中文(繁体) 由 翻译 强力驱动 问题: I want to create a plot like the one below: I know I can use the radarchart function from fmsb package. I wonder if ggplot2 can do so, using polar coordinate? Thanks. 回答1: First, we load some packages. library ( reshape2 ) library ( ggplot2 ) library ( scales ) Here are the data from the radarchart example you linked to. maxmin <- data . frame ( total = c ( 5 , 1 ), phys = c ( 15 , 3 ), psycho = c ( 3 , 0 ), social = c ( 5 , 1 ), env = c ( 5 , 1 ) ) dat <- data . frame ( total = runif ( 3 , 1 , 5 ), phys =

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I have a python script, which I didn't write, that utilises the vtk module. It has worked on my old iMac and on my Linux machines. Trying it on my Macbook Pro with OS X Mavericks installed I get a seg fault during rendering. I have tracked down the error to a call to the vtkRender() method which causes the script to crash with seg fault 11. I literally have no idea how to go further and find out the exact cause of the bug (for instance, how do I step into a method call using pdb?), any ideas? A minimal program crash is as follows: Enthought

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I am new to angular2. I am trying to understand how to use multiple <router-outlets> in a particular template. I have gone though many QA here but couldn't resolve my error. router.module.ts const routes: Routes = [ { path: '', redirectTo: 'one', pathMatch: 'full' }, { path: 'two', component: ClassTwo, children: [ { path: 'three', component: ClassThree, outlet: 'nameThree', }, { path: 'four', component: ClassFour, outlet: 'nameFour' } ] },]; component1.html <h3>In One</h3> <nav> <a routerLink="/two" class="dash-item">...Go to Two...</a> <a