Atlantis Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 16436    Accepted Submission(s): 6706 Problem Description There are several ancient Greek texts that contain descriptions of the fabled i…

这个题算是我的第一个扫描线的题,扫描线算是一种思想吧,用到线段树+离散化.感觉高大上. 主要参考了这位大神的博客. http://www.cnblogs.com/kuangbin/archive/2012/08/15/2640870.html HDU1542 Atlantis(线段树:扫描线) http://acm.hdu.edu.cn/showproblem.php?pid=1542 分析: 首先假设有下图两个矩阵,我们如果用扫描线的方法如何计算它们的总面积呢? 首先我们将矩形的上下边分为上位…

点此看题面 大致题意: 给你\(N\)个矩形,请你求出它们覆盖的面积(重叠的面积只算一次). 扫描线 这道题是一道典型的求矩形面积并问题,是扫描线的一个经典运用.这里就不赘述了. 代码 #include<bits/stdc++.h> #define N 100 using namespace std; int n,cnt,Exist[N<<4]; double Sum[N<<4],xy[(N<<2)+5]; struct Square { int flag,…

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 to…

Atlantis Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9032    Accepted Submission(s): 3873 Problem Description There are several ancient Greek texts that contain descriptions of the fabled i…

题目链接:https://vjudge.net/problem/HDU-1542 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 reg…

英文题面,我就只放个传送门了. Solution  题意是算矩形面积并,这是扫描线算法能解决的经典问题. 算法的大致思想是,把每一个矩形拆成上边和下边(以下称作扫描线),每条扫描线有四个参数l,r,h,v.l和r为它的左右端点的横坐标,h为扫描线的纵坐标,v下面再解释. 然后把扫描线按h从小到大排序,想一想,所有相邻扫描线之间的有效面积(即被矩形覆盖的面积)加起来是不是就是ans? 怎么求呢?我们从下往上处理,设当前处理到第i条扫描线,设第i条扫描线与第i+1条扫描线之间的有效面积为s,那么s=…

题目大意:给n个矩形,可能重叠,求面积. 题目分析:线段树维护扫描线. 代码如下: # include<bits/stdc++.h> using namespace std; # define LL long long # define mid (l+(r-l)/2) const int N=100000; struct Segment { double x1,x2,y; bool buttom; }; Segment seg[(N<<1)+5]; struct Node { do…

题意 给定\(n​\)个矩形\((x_1,y_1,x_2,y_2)​\),求这\(n​\)个矩形的面积并 题解 扫描线裸题,可以不用线段树维护,\(O(n^2)\)是允许的. #include <cstdio> #include <cstring> #include <algorithm> using std::sort; using std::unique; using std::lower_bound; const int N = 1e2 + 10; int n,…

题目链接:点击打开链接 题目描写叙述:给定一些矩形,求这些矩形的总面积.假设有重叠.仅仅算一次 解题思路:扫描线+线段树+离散(代码从上往下扫描) 代码: #include<cstdio> #include <algorithm> #define MAXN 110 #define LL ((rt<<1)+1) #define RR ((rt<<1)+2) using namespace std; int n; struct segment{ double l…