离散化: 将所有的x轴坐标存在一个数组里..排序.当进入一条线段时..通过二分的方式确定其左右点对应的离散值... 扫描线..可以看成一根平行于x轴的直线..至y=0开始往上扫..直到扫出最后一条平行于x轴的边..但是真正在做的时候..不需要完全模拟这个过程..扫描线的做法是从最下面的边开始扫到最上面的边. 线段树: 本题用于动态维护扫描线在往上走时..x哪些区域是有合法面积的.. 几个图说明扫描线扫描..线段树维护的过程..: 初始状态 扫到最下边的线,点更新1~3为1 扫到第二根线,此时将计…

Atlantis Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11777    Accepted Submission(s): 4983 Problem Description There are several ancient Greek texts that contain descriptions of the fabled…

第一次做线段树扫描法的题,网搜各种讲解,发现大多数都讲得太过简洁,不是太容易理解.所以自己打算写一个详细的.看完必会o(∩_∩)o 顾名思义,扫描法就是用一根想象中的线扫过所有矩形,在写代码的过程中,这根线很重要.方向的话,可以左右扫,也可以上下扫.方法是一样的,这里我用的是由下向上的扫描法. 如上图所示,坐标系内有两个矩形.位置分别由左下角和右上角顶点的坐标来给出.上下扫描法是对x轴建立线段树,矩形与y平行的两条边是没有用的,在这里直接去掉.如下图. 现想象有一条线从最下面的边开始依次向上扫描…

这次是求矩形面积并 /* Problem: 1151 User: 96655 Memory: 716K Time: 0MS Language: G++ Result: Accepted */ #include <iostream> #include <algorithm> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include…

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…

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 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 19374   Accepted: 7358 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 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 8998    Accepted Submission(s): 3856 Problem Description There are several ancient Greek texts that contain descriptions of the fabled i…

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…

扫描线终于看懂了...咕咕了快三个月$qwq$ 对于所有的横线按纵坐标排序,矩阵靠下的线权值设为$1$,靠上的线权值设为$-1$,然后执行线段树区间加减,每次的贡献就是有效宽度乘上两次计算时的纵坐标之差. $cnt$数组记录每个位置被覆盖的次数,$sum$数组用来记区间总长度(即有效宽度),所以每一次把$sum[1]$乘上高就行了. 注意到每个$r$都减了$1$,原因是原先的坐标是点的坐标,而现在一个位置代表一个区间. #include<cstdio> #include<iostream…