POJ2482 Stars in Your Window（扫描线+区间最大+区间更新）

Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remember, vividly, on the beautiful Zhuhai Campus, 4 years ago, from the moment I saw you smile, as you were walking out of the classroom and turned…

POJ2482 Stars in Your Window 和 test20180919 区间最大值

Stars in Your Window Language:Default Stars in Your Window Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 15036 Accepted: 4061 Description Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I sti…

poj 2482 Stars in Your Window(扫描线)

id=2482" target="_blank" style="">题目链接:poj 2482 Stars in Your Window 题目大意:平面上有N个星星.问一个W∗H的矩形最多能括进多少个星星. 解题思路:扫描线的变形.仅仅要以每一个点为左上角.建立矩形,这个矩形即为框框左下角放的位置能够括到该点,那么N个星星就有N个矩形,扫描线处理哪些位置覆盖次数最多. #include <cstdio> #include <cstr…

Poj2482 Stars in Your Window（扫描线）

题面 Poj 题解下面内容引用自"李煜东 <算法竞赛进阶指南>"(对原文略有缩减,侵删): 因为矩形的大小固定,所以矩形可以由它的任意一个顶点唯一确定.我们可以考虑把矩形的右上角顶点放在什么位置,圈住的星星亮度总和最大. 所以,对于一颗星星,能够覆盖住这颗星星的右上角的位置在区间\([x,y]-[x+w,y+h]\)之间,但是由于边界上的星星不算,所以不妨将星星变为\([x-0.5,y-0.5]\),于是右上角变为\([x+w-1,y+h-1]\). 问题就转化成了:平面…

POJ2482 Stars in Your Window 题解

poj2482 Stars in Your Window

此题可用线段树或静态二叉树来做. 考虑用线段树: 很容易想到先限定矩形横轴范围再考虑在此纵轴上矩形内物品总价值的最大值. 那么枚举矩形横轴的复杂度是O(n)的,考虑如何快速获取纵轴上的最大值. 我们不可能再次枚举纵轴,依次统计,这样做事多余的. 考虑窗口在纵轴上的滑动,每上升到一个新的高度,在加入新元素的同时只需将最底层的那些值弹出队列即可. 这样我们需要考虑队列上元素和的最大值. 我们从反面考虑每个元素对特定队列(矩形纵轴位置)的贡献. 枚举窗口的上面一条边,那么元素对窗口贡献正值当且仅当H(…

【POJ-2482】Stars in your window 线段树 + 扫描线

Stars in Your Window Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11706 Accepted: 3183 Description Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remember, vividly, on the beaut…

POJ 2482 Stars in Your Window 线段树扫描线

Stars in Your Window Description Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remember, vividly, on the beautiful Zhuhai Campus, 4 years ago, from the moment I saw you smile, as you were walk…

【POJ2482】Stars in Your Window

[POJ2482]Stars in Your Window 题面 vjudge 题解第一眼还真没发现这题居然™是个扫描线令点的坐标为\((x,y)\)权值为\(c\),则若这个点能对结果有\(c\)的贡献,必须要矩形左下角的点的范围必须在\(([x,x+w),[y,y+h))\)之间则按扫描线套路将一个类似矩形的范围拆成线\((x,y1,y2,c)\).\((x+w,y1,y2,-c)\)(依次表示横坐标.下端点纵坐标.上端点纵坐标.权值)即可最后注意排序时不但要按\(x\)排,也要按…

【POJ 2482】 Stars in Your Window（线段树+离散化+扫描线）

[POJ 2482] Stars in Your Window(线段树+离散化+扫描线) Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11294 Accepted: 3091 Description Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remembe…

【POJ2482】【线段树】Stars in Your Window

Description Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remember, vividly, on the beautiful Zhuhai Campus, 4 years ago, from the moment I saw you smile, as you were walking out of the classroo…

poj 2482 Stars in Your Window + 51Nod1208（扫描线+离散化+线段树）

Stars in Your Window Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 13196 Accepted: 3609 Description Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remember, vividly, on the beaut…

51nod 1208 && POJ 2482：Stars in Your Window

1208 Stars in Your Window 题目来源: Poj 基准时间限制:2 秒空间限制:131072 KB 分值: 160 难度:6级算法题收藏取消关注整点上有N颗星星,每颗星星有一个亮度.用一个平行于x轴和y轴,宽为W高为H的方框去套星星.套住的所有星星的亮度之和为S(包括边框上的星星),求S的最大值. Input 第1行:共3个数N, W, H,中间用空格分割,N为星星的数量,W为方框的宽度,H为方框的高度.(2 <= N <= 50000, 1 <= W,…

POJ 2482 Stars in Your Window（线段树）

POJ 2482 Stars in Your Window 题目链接题意:给定一些星星,每一个星星都有一个亮度.如今要用w * h的矩形去框星星,问最大能框的亮度是多少思路:转化为扫描线的问题,每一个星星转化为一个矩形,那么等于求矩形相交区域值最大的区域,利用线段树去维护就可以代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long…

[poj P2482] Stars in Your Window

[poj P2482] Stars in Your Window Time Limit: 1000MS Memory Limit: 65536K Description Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remember, vividly, on the beautiful Zhuhai Campus, 4 years ago…

Stars in Your Window（线段树求最大矩形交）

题目连接 http://poj.org/problem?id=2482 Description Fleeting time does not blur my memory of you. Can it really be 4 years since I first saw you? I still remember, vividly, on the beautiful Zhuhai Campus, 4 years ago, from the moment I saw you smile, as…

(中等) POJ 2482 Stars in Your Window，静态二叉树。

Description Here comes the problem: Assume the sky is a flat plane. All the stars lie on it with a location (x, y). for each star, there is a grade ranging from 1 to 100, representing its brightness, where 100 is the brightest and 1 is the weakest. T…

整数区间及区间集合（C#实现）

/// <summary> /// 整数区间类 /// </summary> private class Interval { , _end = ; public int Start { get { return Math.Min(this._start, this._end); } set { this._start = value; } } public int End { get { return Math.Max(this._start, this._end); } set…

[用CDQ分治解决区间加&区间求和]【习作】

[前言] 作为一个什么数据结构都不会只会CDQ分治和分块的蒟蒻,面对区间加&区间求和这么难的问题,怎么可能会写线段树呢于是,用CDQ分治解决区间加&区间求和这篇习作应运而生 [Part.I]区间加&区间求和的数据结构做法 [一]线段树裸题... 1141ms #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include…

线段树区间开方区间求和 & 区间赋值、加、查询

本文同步发表于 https://www.zybuluo.com/Gary-Ying/note/1288518 线段树的小应用 -- 维护区间开方区间求和题目传送门约定: sum(i,j) 表示区间 [i,j] 中所有元素的和,也就是\(\Sigma_{k=i}^j a_k\) 这个维护思想来自分块 :线段树维护区间开方的难点就在于我们没有办法很方便地维护区间的和,具体来说,如果我们对区间 [l,r] 进行开方,我们并不能从 sum(l,r) 推到 sum'(l,r). 这就比较麻烦了,我们…

树状数组区间加法&区间求和操作

树状数组区间加法&区间求和操作一般的树状数组解决区间加&单点询问并不复杂但是要解决区间求和... 我们假设原数组是\(\{a_i\}\),差分数组\(\{d_i=a_i-a_{i-1}\}\) 所以,我们有式子 \[a_x=\sum_{i=1}^xd_i\] 现在的问题是区间和,也就是求 \[\sum_{i=1}^xa_i\] 如果把每个都拆成上面的形式,那么我们就有 \[Ans=\sum_{i=1}^nd_i(x-i+1)\] \[Ans=\sum_{i=1}^n(x+1)d_i-\…

pandas处理时间序列（2）：DatetimeIndex、索引和选择、含有重复索引的时间序列、日期范围与频率和移位、时间区间和区间算术

一.时间序列基础 1. 时间戳索引DatetimeIndex 生成20个DatetimeIndex from datetime import datetime dates = pd.date_range(start='2019-04-01',periods=20) dates 用这20个索引作为ts的索引 ts = pd.Series(np.random.randn(20),index=dates) ts 不同索引的时间序列之间的算术运算在日期上自动对齐 ts + ts[::2] pandas使…

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