所以可以对点权离散化,然后用线段树来维护上面需要的四个值,即可进行 dsu on tree,整体复杂度 n log n log n n \log n \log n nlognlogn。由于上面的统计我们都是进行的单向计算,所以还要对上述计算完后的答案乘以 2 2 2即可。#include <bits/stdc++.h> #define ls rt <<
tree[rt] = vt[i++];//从vt里读取数据存入tree中 return; }//不是叶节点就走下面的过程 intm = (l+r)>>1; build(l,m,lson); build(m+1,r,rson); pushUp(rt);//每层递归执行到这句,rt的左右儿子肯定是存好了的 } 解释:如果是叶节点,从vt中取数存入。如果不是,分别递归地建左右子树。(...
Segment tree(线段树) 1.线段树的结构和思想 线段树基本结构 简单操作 1.单点修改:时间复杂度O(log n),因为线段树高是O(log n)的,然后会修改这个点到根的路径上的点权,所以是O(log n)的。 2.区间查询(比如:最小值) 实现 #include<bits/stdc++.h> using namespacestd; typedeflonglongll; constintN =...
// printf("%d %d %d %d\n",p,l,r,tree[id].sum[p]); } void modify(int p,int l,int r,int L,int R,int id,int x){ if(L<=l&&r<=R){ tree[id].lazy[p]^=x; tree[id].sum[p]=r-l+1-tree[id].sum[p]; return; } int mid=(l+r)>>1; pushdown(p,id,r-l+1); ...
typedef tree<ll, null_type, less_equal<ll>, rb_tree_tag, tree_order_statistics_node_update> TREE; ll a[MAXN], st[MAXN * 4], lazy[MAXN * 4]; ll F(ll a, ll b) { return a + b; } void build(ll n, ll l, ll r) { lazy[n] = 0; if (l == r) { st[n] = a[...
The segment tree is a highly versatile data structure, based upon the divide-and-conquer paradigm, which can be thought of as a tree of intervals of an underlying array, constructed so that queries on ranges of the array as well as modifications to the array's elements may be efficiently ...
and so on. Porting an old program to the segment model can be easy if one stores the complete program, code, and data in a single segment, but this method loses much of the advantage of the segment model because the entire segment must haveread,write, andexecutepermission. Restructuring an...
Tropical geometry with the max-plus algebra has been applied to statistical learning models over tree spaces because geometry with the tropical metric over tree spaces has some nice properties such as convexity in terms of the tropical metric. One of the challenges in applications of tropical ...
If you are doing zkw segment tree anyways might was well do it in O(n) memory, as opposed to this (arguably more intuitive) O(n^2). Building an O(n^2) segment tree is very expensive, your implementation took O(n^2) just to build the tree. Sparse table can be built in the sam...
分段树(segment tree)是强化学习中 "优先级回放机制" 的重要组成部分。本文针对分段树(segment tree)的一个开源版本的实现来进行分析,代码地址: https://gitee.com/devilmaycry812839668/Rainbow/blob/master/memory.py Transition_dtype = np.dtype( ...