C - Segments POJ - 3304 最近开始刷计算几何了 公式好多完全不会 数学不行 几何不行 记忆力不行 当机 查的题解 就当复习吧 这套专题拿来熟悉一下计算几何模板 #include <iostream> #include<stdio.h> #include<math.h> using namespace std; const double eps = 1e-8; int sgn(double x)//处理精度 { if(fabs(x) < eps…

2019 杭电多校 10 1007 题目链接:HDU 6697 比赛链接:2019 Multi-University Training Contest 10 Problem Description The closest pair of points problem is a well-known problem of computational geometry. In this problem, you are given \(n\) points in the Euclidean plan…

题意: 给出n条线段两个端点的坐标,问所有线段投影到一条直线上,如果这些所有投影至少相交于一点就输出Yes!,否则输出No!. 思路: 计算几何.这道题要思考到两点: 1:把问题转化为是否存在一条直线与每条线段都有交点.证明:若存在一条直线l和所有线段相交,作一条直线m和l垂直,则m就是题中要求的直线,所有线段投影的一个公共点即为垂足. 2:枚举两两线段的各一个端点,连一条直线,再判断剩下的线段是否都和这条直线有交点.证明:若有l和所有线段相交,则可保持l和所有线段相交,左右平移l到和某一线段交…

Segments Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 15335   Accepted: 4862 Description Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments…

Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common. Input Input begins with a number T show…

意甲冠军:给出的一些段的.问:能否找到一条直线,通过所有的行 思维:假设一条直线的存在,所以必须有该过两点的线,然后列举两点,然后推断是否存在与所有的行的交点可以是 代码: #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; struct Point { double x, y; Point() {} Point(dou…

嘟嘟嘟 题面就不说了,网上都有. 刚开始理解成了只要有不孤立的线段就算合法,结果就不会了--然而题中要求是所有线段至少有一个交点. 其实想一想就知道,问题转化为了是否存在一条直线和所有线段都有交点. 所以枚举线段的端点构成直线,然后判断直线和线段是否有交点. 具体做法就是对于直线\(AB\),判断线段\(CD\)是否在\(AB\)的两侧,用叉积即可:如果又向面积符号相同,就说明在一侧,即\((\overrightarrow{AB} \times \overrightarrow{AC}) * (\…

1.poj2318 TOYS 传送:http://poj.org/problem?id=2318 题意:有m个点落在n+1个区域内.问落在每个区域的个数. 分析:二分查找落在哪个区域内.叉积判断点与线段的位置. #include<iostream> #include<cstring> #include<algorithm> using namespace std; ; struct point{ int x,y; point(){ } point(int _x,int…

图论 图论解题报告索引 DFS poj1321 - 棋盘问题 poj1416 - Shredding Company poj2676 - Sudoku poj2488 - A Knight's Journey poj1724 - ROADS(邻接表+DFS) BFS poj3278 - Catch That Cow(空间BFS) poj2251 - Dungeon Master(空间BFS) poj3414 - Pots poj1915 - Knight Moves poj3126 - Prim…

Segments Time Limit: 1000MS Memory Limit: 65536K Description Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least…