Problem Description For a sequence S1, S2, ... , SN, and a pair of integers (i, j), if 1 <= i <= j <= N and Si < Si+1 < Si+2 < ... < Sj-1 < Sj , then the sequence Si, Si+1, ... , Sj is a CIS(Continuous Increasing Subsequence). The…
The LCIS on the Tree Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 175    Accepted Submission(s): 40 Problem Description For a sequence S1, S2, ... , SN, and a pair of integers (i, j), if 1 <=…
Problem Description We have met so many problems on the tree, so today we will have a query problem on a set of trees. There are N nodes, each node will have a unique weight Wi. We will have four kinds of operations on it and you should solve them ef…
2631: tree Time Limit: 30 Sec  Memory Limit: 128 MBSubmit: 5171  Solved: 1754[Submit][Status][Discuss] Description 一棵n个点的树,每个点的初始权值为1.对于这棵树有q个操作,每个操作为以下四种操作之一:+ u v c:将u到v的路径上的点的权值都加上自然数c:- u1 v1 u2 v2:将树中原有的边(u1,v1)删除,加入一条新边(u2,v2),保证操作完之后仍然是一棵树:* u…
GCD Tree Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 415    Accepted Submission(s): 172 Problem Description Teacher Mai has a graph with n vertices numbered from 1 to n. For every edge(u,v),…
Tree Time Limit: 16000/8000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 920    Accepted Submission(s): 388 Problem Description You are given a tree with N nodes which are numbered by integers 1..N. Each node is a…
2631: tree Time Limit: 30 Sec  Memory Limit: 128 MBSubmit: 1716  Solved: 576[Submit][Status] Description 一棵n个点的树,每个点的初始权值为1.对于这棵树有q个操作,每个操作为以下四种操作之一:+ u v c:将u到v的路径上的点的权值都加上自然数c:- u1 v1 u2 v2:将树中原有的边(u1,v1)删除,加入一条新边(u2,v2),保证操作完之后仍然是一棵树:* u v c:将u到v的…
为了优化体验(其实是强迫症),蒟蒻把总结拆成了两篇,方便不同学习阶段的Dalao们切换. LCT总结--应用篇戳这里 概念.性质简述 首先介绍一下链剖分的概念(感谢laofu的讲课) 链剖分,是指一类对树的边进行轻重划分的操作,这样做的目的是为了减少某些链上的修改.查询等操作的复杂度. 目前总共有三类:重链剖分,实链剖分和并不常见的长链剖分 重链剖分 实际上我们经常讲的树剖,就是重链剖分的常用称呼. 对于每个点,选择最大的子树,将这条连边划分为重边,而连向其他子树的边划分为轻边. 若干重边连接在…
题目大意:维护一种树形数据结构.支持下面操作: 1.树上两点之间的点权值+k. 2.删除一条边.添加一条边,保证加边之后还是一棵树. 3.树上两点之间点权值*k. 4.询问树上两点时间点的权值和. 思路:利用动态树维护这棵树,lct的裸题.假设不会下传标记的,先去做BZOJ1798,也是这种标记,仅仅只是在线段树上做,比这个要简单很多. 这个也是我的LCT的第一题,理解起来十分困难啊... CODE: #include <cstdio> #include <cstring> #in…
模板参考:https://blog.csdn.net/saramanda/article/details/55253627 综合各位大大博客后整理的模板: #include<iostream> #include<cstdio> using namespace std; + ; struct LCT { struct node { ]; //父亲(Splay对应的链向上由轻边连着哪个节点).左右儿子 int reverse;//区间反转标记 bool is_root; //是否是所在…