---恢复内容开始--- LINK 题意:同POJ1151 思路: /** @Date : 2017-07-19 13:24:45 * @FileName: POJ 1389 线段树+扫描线+面积并 同1151.cpp * @Platform: Windows * @Author : Lweleth (SoungEarlf@gmail.com) * @Link : https://github.com/ * @Version : $Id$ */ #include <stdio.h> #incl…

请戳此处 #include<cstdio> #include<algorithm> #include<cstring> #define N 1010 #define LEN 60010 using namespace std; struct Edge { int l,r,h,f; bool operator < (const Edge &a) const { if (h==a.h) return f<a.f; return h<a.h; } }…

原题 线段树+扫描线 对于这样一个不规则图形,我们要求他的面积有两种方法,割和补. 补显然不行,因为补完你需要求补上去的内部分不规则图形面积-- 那么怎么割呢? 像这样: 我们就转化成了无数个矩形的和.要想求这些矩形的面积,也就是底成高.因为我们并不懂什么二维线段树,所以我们就用"扫描线"(从左到右的更新计算). 不妨把y轴当做一个线段树,然后维护哪些位置被覆盖了,直到下一条更新,所增加的面积就是被覆盖位置(高)和这两个距离的差(底)的乘积. Eg: 我们第一次更新建了第一条边的树,第…

这个题lba等神犇说可以不用离散化,但是我就是要用. 题干: Description There are N, <= N <= , rectangles -D xy-plane. The four sides of a rectangle are horizontal or vertical line segments. Rectangles are defined by their lower-left and upper-right corner points. Each corner p…

http://poj.org/problem?id=1389 题面描述在二维xy平面中有N,1 <= N <= 1,000个矩形.矩形的四边是水平或垂直线段.矩形由左下角和右上角的点定义.每个角点都是一对两个非负整数,范围从0到50,000,表示其x和y坐标. 求出所有矩形的面积(重叠部分只算一次) 示例:考虑以下三个矩形: 矩形1:<(0,0)(4,4)>, 矩形2:<(1,1)(5,2)>, 矩形3:<(1,1)(2,5)>. 所有由这些矩形构造的简单多…

离散化后,[1,10]=[1,3]+[6,10]就丢了[4,5]这一段了. 因为更新[3,6]时,它只更新到[3,3],[6,6]. 要么在相差大于1的两点间加入一个值,要么就让左右端点为l,r的线段树节点表示到x[l]到x[r+1]的区间. 这样tree[l,r]=tree[l,m]+tree[m+1,r]=[x[l],x[m+1]]+[x[m+1],x[r+1]] #include<iostream> #include<algorithm> #include<cstdio…

poj1389:http://poj.org/problem?id=1389 题意:求矩形面积的并题解:扫描线加线段树 同poj1389 #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; ; int num; struct Node{ int l; int r; int tp; int y; bool operator…

POJ1389 给定n个整数点矩形,求面积并. 显然ans必然是整数. 记录若干个事件,每个矩形的左边的竖边记为开始,右边的竖边记为结束. 进行坐标离散化后用线段树维护每个竖的区间, 就可以快速积分了. #include<stdio.h> #include<stdlib.h> #include<cstring> #include<math.h> #include<algorithm> #include<vector> using na…

Area of Simple Polygons Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 3412   Accepted: 1763 Description There are N, 1 <= N <= 1,000 rectangles in the 2-D xy-plane. The four sides of a rectangle are horizontal or vertical line segment…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5861   Accepted: 2612 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

做这道题之前,建议先做POJ 1151  Atlantis,经典的扫描线求矩阵的面积并 参考连接: http://www.cnblogs.com/scau20110726/archive/2013/04/13/3018702.html 线段树辅助——扫描线法计算矩形周长并(轮廓线):http://www.cnblogs.com/scau20110726/archive/2013/04/13/3018687.htmlhttp://blog.csdn.net/ophunter/article/det…

/* poj 1654 Area 多边形面积 题目意思很简单,但是1000000的point开不了 */ #include<stdio.h> #include<math.h> #include<string.h> const int N=1000000+10; const double eps=1e-8; struct point { double x,y; point(){} point(double a,double b):x(a),y(b){} }; int le…

扫描线算是线段树的一个比较特殊的用法,虽然NOIP不一定会考,但是学学还是有用的,况且也不是很难理解. 以前学过一点,不是很透,今天算是搞懂了. 就以这道题为例吧:嘟嘟嘟 题目的意思是在一个二维坐标系中给了很多矩形,然后求这些矩形的总覆盖面积,也就是面积并. 我就不讲暴力,直接切入正题吧. 扫描线,听这个名字就可以想象一下,现在有这么多重叠的矩形,然后有一个线从下往上扫,那么每一时刻这条线上被覆盖的长度之和,就是我们要求的答案. 那么首先可以想到,要把给定的矩形都离线下来,拆成上下来个面,并标记…

链接:http://poj.org/problem?id=1265 Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4969   Accepted: 2231 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionag…

链接:http://poj.org/problem?id=1654 Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14952   Accepted: 4189 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orth…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17456   Accepted: 4847 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5227   Accepted: 2342 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

Area Time Limit: 1000MS Memory Limit: 10000K Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From this vertex, you may go step by step to the fol…

题目链接:POJ 1265 Problem Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research and development facility the company has installed the latest s…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16894   Accepted: 4698 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

题目链接:http://poj.org/problem?id=1151 题目大意:坐标轴上给你n个矩形, 问这n个矩形覆盖的面积 题目思路:矩形面积并. 代码如下: #include<stdio.h> #include<vector> #include<algorithm> using namespace std; ; struct Line { int l, r, flag; double h; Line(){} Line(int l, int r, int flag…

有一种定理,叫毕克定理....                             Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4352   Accepted: 1977 Description Being well known for its highly innovative products, Merck would definitely be a good target for industria…

题目:http://poj.org/problem?id=1265 题意:已知机器人行走步数及每一步的坐标   变化量 ,求机器人所走路径围成的多边形的面积.多边形边上和内部的点的数量. 思路:1.以格子点为顶点的线段,覆盖的点的个数为GCD(dx,dy),其中,dxdy分别为线段横向占的点数和纵向占的点数.如果dx或dy为0,则覆盖的点数为dy或dx. 2.Pick公式:平面上以格子点为顶点的简单多边形,如果边上的点数为on,内部的点数为in,则它的面积为A=on/2+in-1. 3.任意一个…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 5666   Accepted: 2533 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new resear…

Area Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 4373 Accepted: 1983 Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research a…

id=2482" target="_blank" style="">题目链接:poj 2482 Stars in Your Window 题目大意:平面上有N个星星.问一个W∗H的矩形最多能括进多少个星星. 解题思路:扫描线的变形.仅仅要以每一个点为左上角.建立矩形,这个矩形即为框框左下角放的位置能够括到该点,那么N个星星就有N个矩形,扫描线处理哪些位置覆盖次数最多. #include <cstdio> #include <cstr…

Area Time Limit: 1000MS Memory Limit: 10000K Description Being well known for its highly innovative products, Merck would definitely be a good target for industrial espionage. To protect its brand-new research and development facility the company has…

[题目链接] http://poj.org/problem?id=1981 [题目大意] 给出平面上一些点,问一个半径为1的圆最多可以覆盖几个点 [题解] 我们对于每个点画半径为1的圆,那么在两圆交弧上的点所画的圆,一定可以覆盖这两个点 我们对于每个点计算出其和其它点的交弧,对这些交弧计算起末位置对于圆心的极角, 对这些我们进行扫描线操作,统计最大交集数量就是答案. [代码] #include <cstdio> #include <algorithm> #include <c…

题目:http://poj.org/problem?id=1265 Sample Input 2 4 1 0 0 1 -1 0 0 -1 7 5 0 1 3 -2 2 -1 0 0 -3 -3 1 0 -3 Sample Output Scenario #1: 0 4 1.0 Scenario #2: 12 16 19.0 注意:题目给出的成对的数可不是坐标,是在x和y方向走的数量. 边界上的格点数:一条左开右闭的线段(x1, x2)->(x2, y2)上的格点数为:gcd( abs(x2-x1…

[题目链接] http://poj.org/problem?id=3470 [题目大意] 给出几面墙,均垂直于x轴或者y轴,给出一些鸟的位置(二维坐标点), 鸟只会垂直x轴或者y轴飞行,并且会撞上最近的墙,问每面墙最后会有几只鸟撞上去. [题解] 我们将所有的二维坐标离散,对xy方向分别进行扫描线, 以y轴方向为例,我们先从y最小的线开始扫, 如果是墙,那么在线段树中更新其在x轴上的分布位置 如果是鸟的坐标,那么在线段树中查找其位置,更新其答案. 之后从y最大的开始反向扫描,更新,x方向也同理.…