[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 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23758   Accepted: 8834 Description There are several ancient Greek texts that contain descriptions of the fabled island Atlantis. Some of these texts even include maps of parts…

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. Your friend…

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…

题目链接:http://poj.org/problem?id=1151 扫描线+离散+线段树,线段树每个节点保存的是离散后节点右边的线段. //STATUS:C++_AC_16MS_208KB #include <functional> #include <algorithm> #include <iostream> //#include <ext/rope> #include <fstream> #include <sstream>…

有点难,扫描线易懂,离散化然后线段树处理有点不太好理解. 因为这里是一个区间,所有在线段树中更新时,必须是一个长度大于1的区间才是有效的,比如[l,l]这是一根线段,而不是区间了. AC代码 #include <stdio.h> #include <map> #include <vector> #include <algorithm> using namespace std; + ; struct Line{ double x, y1, y2; int fl…

http://poj.org/problem?id=1151 想学一下扫描线线段树,结果写了道水题. #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #include<cmath> #include<set> using namespace std; ; int n; ]; ]; ],c[maxn*]; ,tot1=; int main…

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…

题目:http://poj.org/problem?id=1151 经典的扫描线问题: 可以用线段树的每个点代表横向被矩形上下边分割开的每一格,这样将一个矩形的出现或消失化为线段树上的单点修改: 每个格子记录两个值:c(矩形存在情况),sum(对当前答案作出贡献的长度): 将y离散化作为建树的依据: 一开始没想到线段树上的点应该是横向的格子,写了个乱七八糟: #include<iostream> #include<cstdio> #include<cstring> #i…

题目大意 给出几个矩形对角端点坐标,求这些矩形整体覆盖的面积. 扫描线算法 整个平面被每个矩形的水平边所在直线(以后简称“水平线”)分成了几个部分,而整体覆盖面积则为每相邻的两个水平线间夹的长度(以后简称“夹长”)以及水平线所对应的一个或几个矩形水平边(以后简称“水平线段”)在水平线中所占的长度之积的和.于是我们假设有一个扫描线从下往上扫.由排序得到夹长很容易,但是水平线段长就要用线段树了. 线段树 节点维护的是什么 水平线段在两个点[l,r]之间所占的长度CoverLen. l,r是doubl…