也就是说对于修改操作,实际上是每减小一个势能用了O(logn)O(logn)的代价。 考虑修改操作,每次只会修改O(logn)O(logn)节点,最多使势能增加O(logn)O(logn)所以总复杂度是O(nlog2n)O(nlog2n)。 code: Codeforces 1290 E /*program by mangoyang*/#pragmaGCC optimize("Ofast","...
The only variant that stands out a bit is the segment tree beats, because unlike other variants of the segment tree its running time is amortized. Maybe a name like "amortized segment tree" would have been more sound than the current one. At least, it would reflect the main difference ...
浅谈近期用到的segment tree beats technology Segment Tree beatsSegment Tree beats是势能线段树的全称,泛指这种在某些情况下打tag,在某些情况暴力子树,复杂度通过势能分析证明的线段树。 主要用途:区间取min/maxmin/max, 区间开根号 , 区间除xx取整 之类不易打tag处理的区间操作。
[Segment tree Beats! || 分块] Codeforces 793F Tinkoff Challenge - Elimination Round F. Julia the snail 我们用fi表示左边界是i的答案 一条线段[a,b]对答案的影响是 f1到fl中大于等于a的都跟b取max 这个可以用线段树科技做 复杂度证明跟区间最值操作的势能分析应该差不多...
Cancel Submit feedback Saved searches Use saved searches to filter your results more quickly Cancel Create saved search Sign in Sign up Reseting focus {{ message }} EndlessCheng / codeforces-go Public Notifications You must be signed in to change notification settings Fork...
Cancel Submit feedback Saved searches Use saved searches to filter your results more quickly Cancel Create saved search Sign in Sign up Reseting focus {{ message }} EndlessCheng / codeforces-go Public Notifications You must be signed in to change notification settings F...
Segment Tree Beats are a really cool, and fairly advanced, modification you can make to segment trees in order to handle range-min-with queries. I talk about what they are and why they work inEpisode 4 of AlgorithmsThread. Feel free to comment any questions about what I covered in the ...
Segment Tree Beats 1 给出 nnn 个数,qqq 次询问,求最大子段和,相同的数只算一次。(GSS2) 在线不是很好做,我们离线下来,把询问按 rrr 升序排序,线段树上维护每个 lll 到 rrr 的和。如果预处理出每个数上一个出现的位置 pre[i]pre[i]pre[i],那么每次右端点右移相当于区间 [pre[i]+1,r][pre[i...
所以我们使用 Segment Tree Beats 维护区间 minmin、区间加、全局和即可,O(nlog2n)O(nlog2n)。#include<iostream> #define rep(i,a,b) for(int i = (a); i <= (b); i++) #define per(i,b,a) for(int i = (b); i >= (a); i--) #define N 160000 #define Inf 0x3f3f3f3f...
Cancel Submit feedback Saved searches Use saved searches to filter your results more quickly Cancel Create saved search Sign in Sign up Reseting focus {{ message }} EndlessCheng / codeforces-go Public Notifications You must be signed in to change notification settings For...