// 线段和矩形相交 POJ 1410 // #include <bits/stdc++.h> #include <iostream> #include <cstdio> #include <cstdlib> #include <algorithm> #include <vector> #include <math.h> using namespace std; #define LL long long typedef p

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

Atlantis Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12059    Accepted Submission(s): 5083 Problem Description There are several ancient Greek texts that contain descriptions of the fabled i

Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16322   Accepted: 4213 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point: (4,9)

题目链接 Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12040   Accepted: 3125 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point:

矩形相交 包含 问题.参考 假定矩形是用一对点表达的(minx, miny) (maxx, maxy),那么两个矩形    rect1{(minx1, miny1)(maxx1, maxy1)}    rect2{(minx2, miny2)(maxx2, maxy2)}  相交的结果一定是个矩形,构成这个相交矩形rect{(minx, miny) (maxx, maxy)}的点对坐标是:      minx   =   max(minx1,   minx2)      miny   =   m

链接 多边形的面积求解是通过选取一个点(通常为原点或者多边形的第一个点)和其它边组成的三角形的有向面积. 对于两个多边形的相交面积就可以通过把多边形分解为三角形,求出三角形的有向面积递加.三角形为凸多边形,因此可以直接用凸多边形相交求面积的模板. 凸多边形相交后的部分肯定还是凸多边形,所以只需要判断哪些点是相交部分上的点,最后求下面积. #include <iostream> #include<cstdio> #include<cstring> #include<

Matt is a big fan of logo design. Recently he falls in love with logo made up by rings. The following figures are some famous examples you may know. A ring is a 2-D figure bounded by two circles sharing the common center. The radius for these circles

J. Polygons Intersection time limit per test:2 seconds memory limit per test:64 megabytes input:standard input output:standard output We will not waste your time, it is a straightforward problem. Given multiple polygons, calculate the area of their i

描述 The Archeologists of the Current Millenium (ACM) now and then discover ancient artifacts located at vertices of regular polygons. The moving sand dunes of the desert render the excavations difficult and thus once three vertices of a polygon are di

C. Masha and two friends time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Recently, Masha was presented with a chessboard with a height of nn and a width of mm. The rows on the chessboard ar

Description The city of M is a famous shopping city and its open-air shopping malls are extremely attractive. During the tourist seasons, thousands of people crowded into these shopping malls and enjoy the vary-different shopping. Unfortunately, the

最近需要用到矩形相交算法的简单应用,所以特地拿一个很简单的算法出来供新手参考,为什么说是给新手的参考呢因为这个算法效率并不是很高,但是这个算法只有简简单单的三行.程序使用了两种方法来判断是否重叠/相交,如果有兴趣可以看一下,如果觉得有bug可以留言.代码仅供参考. C#中矩形的方法为Rectangl(起始点坐标, 矩形的大小)或Rectangl(起始点x坐标, 起始点y坐标, 矩形宽, 矩形高),起始点为矩形区域的左上角. 方法一 姑且叫做“井字法”吧,延长其中一个矩形的四边使其形成一“井”字(

题目传送门 Hard problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1066    Accepted Submission(s): 622 Problem Description cjj is fun with math problem. One day he found a Olympic Mathematics pr

Open-air shopping malls Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2458    Accepted Submission(s): 906 Problem Description The city of M is a famous shopping city and its open-air shopping

题目链接:http://poj.org/problem?id=2082 把矩形按照高度一次递增的循序排列,当违反这一规则的时候,更新ans,用新的data替换之前的矩形.然后最后扫一遍. #include <iostream> #include <stack> #include <cstdio> using namespace std; struct rec { int w; ///宽 int h; ///高 } data; int main() { int n; )

Problem Description Given two rectangles and the coordinates of two points on the diagonals of each rectangle,you have to calculate the area of the intersected part of two rectangles. its sides are parallel to OX and OY .   Input Input The first line

题目: Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point: (4,9) end point: (11,2) rectangle: left-top: (1,5) right-bottom: (7,1)  Figure 1: Line segment doe

题目原网址:http://poj.org/problem?id=1177 题目中文翻译: 解题思路: 总体思路: 1.沿X轴离散化建树 2.按Y值从小到大排序平行与X轴的边,然后顺序处理 如果遇到矩形下面那条边则插入到线段树中,遇到矩形上面的边则将相应的边删除掉 根据线段树当前的状态统计长度 第二点是本题的核心思想,偶再举个例:   第一次求出的部分很好理解. 第二次求出的为什么会少了中间那部分.那是因为插入的新线段覆盖了第一条,此时线段树返回的长度是新的那一条的长度,将这个值再减去上次的就少了

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