扫描线算法,推荐还是用标准的模板去写,treemap只适合于求最大的overlap个数的题目,其余的不能用treemap来解,所以推荐还是用event的思想去+1, -1然后排序扫描的方法可以用来解所有的题型; Number of Airplanes in the Sky 思路:经典扫描线算法:把interval起飞和降落做为event,全部打散,按照时间排列,同时时间相等的,按照降落在前面,起
覆盖的面积 Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4417 Accepted Submission(s): 2180 Problem Description 给定平面上若干矩形,求出被这些矩形覆
Atlantis Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 16755 Accepted: 6391 Description There are several ancient Greek texts that contain descript
题目链接:247. 亚特兰蒂斯 算法分析 只说细节。 这道题目不是特别严格的线段树,因为线段树维护的信息具有区间可合并性,这道题目想了很久想不出可以区间直接合并的信息。我们在这里维护的是离散化后表示小段的数组,设为 a [ ] a[] a[]吧,对于一个边界的两个纵坐标,设为 y 1 y1 y1和 y 2 y2 y2,离散化后的结果为 [ a l , a r ] [al,ar] [al,ar
There are n pirate chests buried in Byteland, labeled by 1,2,…,n. The i-th chest’s location is (xi,yi), and its value is wi, wi can be negative since the pirate can add some poisonous gases into the c
//魔改线段树的做法,想了好久注意是那个一次用len记录,两次用length来计数#include<iostream>#include<cstdio>#include<string.h>#include<algorithm>using namespace std;const int maxn=2*1010;struct tree{double l,r,len,length;int l