题意:给n个可能相交的矩形,问你不重复的总面积 思路:扫描线,一边扫一边加. 扫描线:图片来源:理解扫描线 假设我们要算以下四个矩形面积,显然中间深色的是重复的.我们按照x的大小,从左往右扫,然后用线段树维护y轴向的长度就可以了.但是,我们不能用点去维护y轴坐标,而是抽象成把点i看成y[i]到y[i+1]这个区间,不然会有错. 代码: #include<set> #include<map> #include<stack> #include<cmath> #i

[POJ1151]Atlantis 试题描述 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 Atlantis.

[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

[POJ1151]Atlantis(线段树,扫描线) 题面 Vjudge 题解 学一学扫描线 其实很简单啦 这道题目要求的就是若干矩形的面积和 把扫描线平行于某个轴扫过去(我选的平行\(y\)轴扫) 这样只需要求出每次和\(x\)轴覆盖的长度 就可以两两相乘,求出面积 最后累计和就行啦 #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cma

n个矩形,可以重叠,求面积并. n<=100: 暴力模拟扫描线.模拟赛大水题.(n^2) 甚至网上一种“分块”:分成n^2块,每一块看是否属于一个矩形. 甚至这个题就可以这么做. n<=100000 这就要到扫描线了. 之前一直不会. 不好处理的地方在于:不知道区间被覆盖怎么计算.... 像sum区间和一样计算??sum>区间长度也不一定完全包含... 不管覆盖多少次,就只算一次??之后的减法怎么算?? 总之, 这篇题解打消了我的疑惑:ACM POJ1151 (HDU 1542) Atl

 描述 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 Atlantis. Your friend Bill ha

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1542 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Problem Description There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. So

Atlantis 题目连接 http://poj.org/problem?id=1151 Description here 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 describ

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 Atlantis. You

Atlantis Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16882   Accepted: 6435 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

Atlantis poj1151 线段树扫描线 题意 题目给了n个矩形,每个矩形给了左下角和右上角的坐标,矩形可能会重叠,求的是矩形最后的面积. 题解思路 这个是我线段树扫描线的第一题,听了学长的讲解,仔细研读了学长的代码,也算初步入门. 这里我们竖着的扫描线,从左到右来进行扫描的. 线段树这里端点代表的是一个区间 这里的y范围比较大,所以咱们需要离散化. 代码实现 #include<cstdio> #include<cstring> #include<algorithm&g

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11558 Accepted Submission(s): 4910 Problem Description There are several ancient Greek texts that contain descriptions of the fabled island Atlantis

  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 Atlantis. Your friend Bill has

扫描线求周长: hdu1828 Picture(线段树+扫描线+矩形周长) 参考链接:https://blog.csdn.net/konghhhhh/java/article/details/78236036 假想有一条扫描线,从左往右(从右往左),或者从下往上(从上往下)扫描过整个多边形(或者说畸形..多个矩形叠加后的那个图形).如果是竖直方向上扫描,则是离散化横坐标,如果是水平方向上扫描,则是离散化纵坐标.下面的分析都是离散化横坐标的,并且从下往上扫描的. 扫描之前还需要做一个工作,就是保存

我要举报本次校赛出题人的消极出题!!! 官方题解请戳:http://3.scnuacm2015.sinaapp.com/?p=89(其实就是一堆代码没有题解) A. 树链剖分数据结构板题 题目大意:我没看,看不懂. 基本思路:我不会. 参考代码:找Oyk老师和Czj老师去. B. The background of water problem 题目大意(大写加粗的水题):给定$N$个学生和他们$K$个科目的成绩$S_i$,再给出各科目$K_i$的权重顺序$Q_i$,求排名之后,拥有id为$X$的

以下代码为了阅读方便,省去以下头文件: #include <iostream> #include <stdio.h> #include <math.h> #include <string.h> #include <time.h> #include <stdlib.h> #include <string> #include <bitset> #include <vector> #include <

2561: 最小生成树(题解) Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 1628  Solved: 786 传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=2561 Description 给定一个边带正权的连通无向图G=(V,E),其中N=|V|,M=|E|,N个点从1到N依次编号,给定三个正整数u,v,和L (u≠v),假设现在加入一条边权为L的边(u,v),那么需要删掉最少多少条

Problems     # Name     A Infinite Sequence standard input/output 1 s, 256 MB    x3509 B Restoring Painting standard input/output 1 s, 256 MB    x2519 C Money Transfers standard input/output 1 s, 256 MB    x724 D Tree Construction standard input/outp

题目链接 提交连接:http://acm-software.hrbust.edu.cn/problemset.php?page=5 1470-1482 只做出来四道比较水的题目,还需要加强中等题的训练. 题解: E666 这个题是让求有多少个子串只含有6.寻找连续的6,然后用n*(n+1)/2求出这一段的子串个数,然后把每一段连续的加起来. 做的时候wa了很多次,原来是在n*(n+1)的地方已经超过int型了,所以需要设置类型为long long. #include <cstdio> #inc

A.Relic Discovery_hdu5982 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 57    Accepted Submission(s): 49 Problem Description Recently, paleoanthropologists have found historical remains on an