算法 线段树 + 离散化 思路 对\((x,y,h)\)的左右端点\(x,y\)进行离散化,离散化前的原值记为\(val[i]\),对每个矩形按高度\(h\)从小到大排序. 设离散化后的端点有\(M\)个,则对如图所示\(M-1\)个规则矩形编号为\([1,M-1]\),可以由\(h_{[i, i+1]}\times(val[i+1] - val[i])\)得出第\(i\)个矩形的面积. 开一颗区间为\([1,M-1]\)的线段树,按\(h\)从小到大依次对线段树区间覆盖,可以保证高的矩形覆盖了…

P2061 [USACO07OPEN]城市的地平线City Horizon 扫描线 扫描线简化版 流程(本题为例): 把一个矩形用两条线段(底端点的坐标,向上长度,添加$or$删除)表示,按横坐标排序 $upd:$本题的底端点坐标简化为$(x,0)$ 蓝后对纵坐标建一棵线段树(本题需要对高度进行离散化). 每次对线段树进行覆盖$or$删除区间操作,顺便统计一下$k=$有多少点被覆盖到 而两次(线段)操作之间的长度为$r=x_{i}-x_{i-1}$ 于是两条线段之间被覆盖的面积即为$k*r$ (…

简化版的矩形面积并,不用线段树,不用离散化,代码意外的简单 扫描线,这里的基本思路就是把要求的图形竖着切几刀分成许多矩形,求面积并.(切法就是每出现一条与y轴平行的线段都切一刀) 对于每一个切出来的矩形在处理其右边的线段时计算面积的贡献, #include<cstdio> #include<algorithm> #include<set> using namespace std; typedef long long LL; struct Q { int a,b,h; }…

题目描述 Farmer John has taken his cows on a trip to the city! As the sun sets, the cows gaze at the city horizon and observe the beautiful silhouettes formed by the rectangular buildings. The entire horizon is represented by a number line with N (1 ≤ N…

Description 约翰带着奶牛去都市观光.在落日的余晖里,他们看到了一幢接一幢的摩天高楼的轮廓在地平线 上形成美丽的图案.以地平线为 X 轴,每幢高楼的轮廓是一个位于地平线上的矩形,彼此间可能有 重叠的部分.奶牛一共看到了 N 幢高楼,第 i 幢楼的高度是 Hi,两条边界轮廓在地平线上的坐标是 Ai 到 Bi.请帮助奶牛们计算一下,所有摩天高楼的轮廓覆盖的总面积是多少. Input 第一行一个整数N,然后有N行,每行三个正整数ai.bi.Hi. Output 一个数,数列中所有元素的和.…

1645: [Usaco2007 Open]City Horizon 城市地平线 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 315  Solved: 157[Submit][Status] Description Farmer John has taken his cows on a trip to the city! As the sun sets, the cows gaze at the city horizon and observe t…

BZOJ_1654_[Usaco2007 Open]City Horizon 城市地平线_扫描线 Description N个矩形块,交求面积并. Input * Line 1: A single integer: N * Lines 2..N+1: Input line i+1 describes building i with three space-separated integers: A_i, B_i, and H_i Output * Line 1: The total area,…

[BZOJ1645][Usaco2007 Open]City Horizon 城市地平线 Description Farmer John has taken his cows on a trip to the city! As the sun sets, the cows gaze at the city horizon and observe the beautiful silhouettes formed by the rectangular buildings. The entire ho…

Description Farmer John has taken his cows on a trip to the city! As the sun sets, the cows gaze at the city horizon and observe the beautiful silhouettes formed by the rectangular buildings. The entire horizon is represented by a number line with N…

http://www.lydsy.com/JudgeOnline/problem.php?id=1645 这题的方法很奇妙啊...一开始我打了一个“离散”后的线段树.............果然爆了..(因为压根没离散) 这题我们可以画图知道,每2个点都有一个区间,而这个区间的高度是一样的,因此,我们只需要找相邻的两个点,用他们的距离×这个区间的高度就是这块矩形的面积. 将所有这样的矩形累计起来就是答案了. 因此线段树就离散到了O(n)的大小..真神.. 只需要维护点的位置,然后维护区间最值即可…