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…

---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…

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 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…

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…

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; } }…

题目链接:点击打开链接 题目描写叙述:给定一些矩形,求这些矩形的总面积.假设有重叠.仅仅算一次 解题思路:扫描线+线段树+离散(代码从上往下扫描) 代码: #include<cstdio> #include <algorithm> #define MAXN 110 #define LL ((rt<<1)+1) #define RR ((rt<<1)+2) using namespace std; int n; struct segment{ double l…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3642 Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description Jack knows that there is a great underground treasury in a secret region. And he has a special d…

Description Ted has a new house with a huge window. In this big summer, Ted decides to decorate the window with some posters to prevent the glare outside. All things that Ted can find are rectangle posters. However, Ted is such a picky guy that in ev…

[BZOJ4999]This Problem Is Too Simple!(线段树) 题面 BZOJ 题解 对于每个值,维护一棵线段树就好啦 动态开点,否则空间开不下 剩下的就是很简单的问题啦 当然了,对于数值要离散化 没必要离线吧,在线用\(map\)维护就行了 #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #inc…