zoukankan      html  css  js  c++  java
  • Balanced Playlist from Codeforces Global Round 5

    https://codeforces.com/contest/1237/problem/D

    i want to remind of the diversity of solutions to this facinating problem.

    description

    you have a playlist consists of (n) CDs. the playlist is automatic and cyclic, which means once you choose to start with (i)-th CD, whenever it is finished, CD number (i+1) (if (i<n)) or (1) (if (i=n)) will start playing automatically.

    each CD is assigned a coolness as you like. once you start music, at any moment, you will remember the maximum coolness (x) you have ever heard. you party will terminate the moment you are going to go through a CD with coolness less than (x/2).

    for every CD you play initially, find how many CDs you will listen before you turn off the music.

    breakthough

    update the second loop

    first we have to generate an idea of brute force in O((n^2)), then we can try optimization.

    when the answer of (i)-th CD is calculated, according to our brute Force we have found out the terminal and we rename it as (j). which leads to (ans=j-i). meanwhile, we denote (f[i]=j) as the terminal of (i)-th CD

    note that we have recalculate something in the second loop, which means that between (i) and (i+1) some iterations of (j) overlap. we need to clarify how it happens.

    denote (max) as the maximum coolness of CDs between (i) and (j), denote (maxPos) as where (max) locates. then it occurs to us that for any (k) from (i) to (maxPos), (f[k]=f[i]=j). that is the redundancy and an oppotunity for us to improve the following:

    • just skip this or directly jump from (i) to (max+1) in the first loop.

    what about (k) from (max+1) to (j)? since the limitation of (j) is removed the moment (max) is out, it is obvious that (f[k]>f[i]=j) . here is the other improvement:

    • let the pointer (j) in the second loop stays there instead of refreshing it to be (i+1)

    implement both and you will get this problem passed.

    to calculate the max element of a segment, use segment tree or RMQ, or
    data structure of treap like map or priority queue

  • 相关阅读:
    【MySQL 5.7 Reference Manual】15.4.2 Change Buffer(变更缓冲)
    从MySQL slave system lock延迟说开去
    一文搞懂Raft算法
    Relinking Oracle Home FAQ ( Frequently Asked Questions) (Doc ID 1467060.1)
    Oracle 单实例 Relink Binary Options 说明
    oracle upgrade best pratics
    sysbench安装、使用、结果解读
    Linux 进程管理之四大名捕
    iOS 严重问题解释(crash)
    iOS 图片剪切和压缩的几个方法
  • 原文地址:https://www.cnblogs.com/reshuffle/p/12547385.html
Copyright © 2011-2022 走看看