Atlantis Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10208    Accepted Submission(s): 4351 Problem Description There are several ancient Greek texts that contain descriptions of the fabled i…

学习扫描线ing... 玄学的东西... 扫描线其实就是用一条假想的线去扫描一堆矩形,借以求出他们的面积或周长(这一篇是面积,下一篇是周长) 扫描线求面积的主要思想就是对一个二维的矩形的某一维上建立一棵线段树,然后把另一维按高度排序,从下向上枚举即可. 主题思想其他博客说的很明白了,这里重点记录一下细节问题: 下面认为对横坐标建立线段树扫描纵坐标: 首先,由于读入的都是浮点数,所以我们需要对这个东西离散化,具体做法是先去重再进行二分查找,以下标代替浮点数值. 其次,由于普通线段树维护的是一个散点…

点击打开链接 题意:给你n个矩形,求它们的面积,反复的不反复计算 思路:用线段树的扫描线完毕.将X坐标离散化后,从下到上扫描矩形,进行各种处理,看代码凝视把 #include <stdio.h> #include <string.h> #include <stdlib.h> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; con…

//永远只考虑根节点的信息,说明在query时不会调用pushdown //所有操作均是成对出现,且先加后减 // #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <vector> using namespace std; ; int n; //存每一个操作 struct Segment { //区间长度 double x…

也是很久之前的题目,一直没做 做完之后觉得基本的离散化和扫描线还是不难的,由于本题要离散x点的坐标,最后要计算被覆盖的x轴上的长度,所以不能用普通的建树法,建树建到r-l==1的时候就停止,表示某段而不是某点,同样,左子树和右子树要变成 L MID , MID R 比如1-4子树就是 1-2,2-4...2-4再分成2-3,3-4. 然后就是经典的扫描线用法,对下边设标记为1,上边设标记为-1,每次求得x轴被覆盖的长度,乘以和下一条线段的距离(即矩形的高)即可 #include <iostrea…

Atlantis Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8327    Accepted Submission(s): 3627 Problem Description There are several ancient Greek texts that contain descriptions of the fabled is…

Atlantis Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5932    Accepted Submission(s): 2599 Problem Description There are several ancient Greek texts that contain descriptions of the fabled is…

Picture Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3897    Accepted Submission(s): 1978 Problem Description A number of rectangular posters, photographs and other pictures of the same shape…

Adding New Machine Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1428    Accepted Submission(s): 298 Problem Description Incredible Crazily Progressing Company (ICPC) suffered a lot with the…

威威猫系列故事——晒被子 Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 1592    Accepted Submission(s): 444 Problem Description 因为马拉松初赛中吃鸡腿的题目让不少人抱憾而归,威威猫一直觉得愧对大家,这几天他悄悄搬到直角坐标系里去住了. 生活还要继续,太阳也照常升起,今天,威威猫在…

Weak Pair Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 439    Accepted Submission(s): 155 Problem Description You are given a rooted tree of N nodes, labeled from 1 to N. To the ith node a…

Assign the task Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1647    Accepted Submission(s): 753 Problem Description There is a company that has N employees(numbered from 1 to N),every emplo…

Queue-jumpers Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3348    Accepted Submission(s): 904 Problem Description Ponyo and Garfield are waiting outside the box-office for their favorite mo…

Sequence operation Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7502    Accepted Submission(s): 2233 Problem Description lxhgww got a sequence contains n characters which are all '0's or '1…

Transformation Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 65535/65536 K (Java/Others) Total Submission(s): 4095    Accepted Submission(s): 1008 Problem Description Yuanfang is puzzled with the question below:  There are n integers, a1,…

Memory Control Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5913    Accepted Submission(s): 1380 Problem Description Memory units are numbered from 1 up to N. A sequence of memory units is c…

A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4996    Accepted Submission(s): 1576 Problem Description Let A1, A2, ... , AN be N elements. You need to deal with…

Mex Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 2482    Accepted Submission(s): 805 Problem Description Mex is a function on a set of integers, which is universally used for impartial game…

Level up Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3973    Accepted Submission(s): 1104 Problem Description Level up is the task of all online games. It's very boooooooooring. There is o…

I Hate It Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 57279    Accepted Submission(s): 22365 Problem Description 很多学校流行一种比较的习惯.老师们很喜欢询问,从某某到某某当中,分数最高的是多少. 这让很多学生很反感. 不管你喜不喜欢,现在需要你做的是,就是按照老师…

敌兵布阵 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 66634    Accepted Submission(s): 28074 Problem Description C国的死对头A国这段时间正在进行军事演习,所以C国间谍头子Derek和他手下Tidy又开始忙乎了.A国在海岸线沿直线布置了N个工兵营地,Derek和Tidy的任务…

Can you answer these queries? HDU 4027 线段树 题意 是说有从1到编号的船,每个船都有自己战斗值,然后我方有一个秘密武器,可以使得从一段编号内的船的战斗值变为原来值开根号下的值.有两种操作,第一种就是上面描述的那种,第二种就是询问某个区间内的船的战斗值的总和. 解题思路 使用线段树就不用多说了,关键是如果不优化的话会超时,因为每次修改都是需要递归到叶子节点,很麻烦,但是我们发现,如果一个叶子节点的值已经是1的话,那个再开方它也是1,不变,这样我们就只需要判断…

敌兵布阵 HDU 1166 线段树 题意 这个题是用中文来描写的,很简单,没什么弯. 解题思路 这个题肯定就是用线段树来做了,不过当时想了一下可不可用差分来做,因为不熟练就还是用了线段树来做,几乎就是模板题了. 代码实现 #include<cstdio> #include<cstring> #include<algorithm> #include<string> #include<iostream> # define ls (rt<<…

[题目] Atlantis Problem Description There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts of the island. But unfortunately, these maps describe different regions of…

https://cn.vjudge.net/problem/HDU-1542 题意 求矩形的面积并 分析 点为浮点数,需要离散化处理. 给定一个矩形的左下角坐标和右上角坐标分别为:(x1,y1).(x2,y2),对这样的一个矩形,我们构造两条线段,一条定位在x1,它在y坐标的区间是[y1,y2],并且给定一个cover域值为1:另一条线段定位在x2,区间一样是[y1,y2],给定它一个cover值为-1.根据这样的方法对每个矩形都构造两个线段,最后将所有的线段根据所定位的x从左到右进行排序.插入…

题目链接 题意 给出n个矩形,求面积并. 思路 使用扫描线,我这里离散化y轴,按照x坐标从左往右扫过去.离散化后的y轴可以用线段树维护整个y上面的线段总长度,当碰到扫描线的时候,就可以统计面积.这里要注意线段树上结点维护的是线段的信息,而不是点的信息. 参考资料 #include <bits/stdc++.h> using namespace std; typedef long long LL; typedef pair<int, int> pii; const int INF =…

http://acm.hdu.edu.cn/showproblem.php?pid=1255 覆盖的面积 Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2933    Accepted Submission(s): 1447 Problem Description 给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积.   …

http://acm.hdu.edu.cn/showproblem.php?pid=1542 我的做法是把x轴的表示为线段,然后更新y 不考虑什么优化的话,开始的时候,把他们表达成线段,并按y排序,然后第一次加入线段树的应该就是最底下那条,然后第二条的时候,我们可以询问第二条那段区间,有多少是已经被覆盖的,然后把面积算上就可以. 所以如果区间都是整数,而且数值很少,那么就是线段树成段覆盖的问题了.但是这里是浮点数而且很大. 所以只能把它离散化. 这个时候线段树就不是连续的了,这里就有bug,问题…

http://acm.hdu.edu.cn/showproblem.php?pid=1255 典型线段树辅助扫描线,顾名思义扫描线就是相当于yy出一条直线从左到右(也可以从上到下)扫描过去,此时先将所有的横坐标和纵坐标排序 因为是从左到右扫描,那么横坐标应该离散化一下 当扫描线依次扫描的时候,依次扫描到的纵区间在线段树中查找,依据是上边还是下边记录,上边就是-1,下边就是+1, 如果某区间记录值为0的时候,代表没有被覆盖,为1的时候代表覆盖一次,为2代表覆盖两次(不会出现为负数的情况) 最后将依…

覆盖的面积 Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Problem Description 给定平面上若干矩形,求出被这些矩形覆盖过至少两次的区域的面积.   Input 输入数据的第一行是一个正整数T(1<=T<=100),代表测试数据的数量.每个测试数据的第一行是一个正整数N(1<=N<=1000),代表矩形的数量,然后是N行数据,每一行包含四个浮点…