Intersecting Lines Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8637   Accepted: 3915 Description We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three…

题目传送门:POJ 1269 Intersecting Lines Description We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in…

id=1269" rel="nofollow">Intersecting Lines 大意:给你两条直线的坐标,推断两条直线是否共线.平行.相交.若相交.求出交点. 思路:线段相交推断.求交点的水题.没什么好说的. struct Point{ double x, y; } ; struct Line{ Point a, b; } A, B; double xmult(Point p1, Point p2, Point p) { return (p1.x-p.x)*(p2…

题目链接:http://poj.org/problem?id=1269 Time Limit: 1000MS Memory Limit: 10000K Description We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection b…

题意:给两条直线,判断相交,重合或者平行 思路:判断重合可以用叉积,平行用斜率,其他情况即为相交. 求交点: 这里也用到叉积的原理.假设交点为p0(x0,y0).则有: (p1-p0)X(p2-p0)=0 (p3-p0)X(p2-p0)=0 展开后即是 (y1-y2)x0+(x2-x1)y0+x1y2-x2y1=0 (y3-y4)x0+(x4-x3)y0+x3y4-x4y3=0 将x0,y0作为变量求解二元一次方程组. 假设有二元一次方程组 a1x+b1y+c1=0; a2x+b2y+c2=0…

Intersecting Lines Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8342   Accepted: 3789 Description We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three…

题意:    判断直线间位置关系: 相交,平行,重合 include <iostream> #include <cstdio> using namespace std; struct Point { int x , y; Point(, ) :x(a), y(b) {} }; struct Line { Point s, e; int a, b, c;//a>=0 Line() {} Line(Point s1,Point e1) : s(s1), e(e1) {} void…

题目:http://poj.org/problem?id=1269 相关知识: 叉积求面积:https://www.cnblogs.com/xiexinxinlove/p/3708147.html什么是叉积:https://blog.csdn.net/sunbobosun56801/article/details/78980467        其二维:https://blog.csdn.net/qq_38182397/article/details/80508303计算交点:    方法1:面…

题目链接:http://poj.org/problem?id=1269 题目大意:给出四个点的坐标x1,y1,x2,y2,x3,y3,x4,y4,前两个形成一条直线,后两个坐标形成一条直线.然后问你是否平行,重叠或者相交,如果相交,求出交点坐标. 算法:二维几何直线相交+叉积 解法:先用叉积判断是否相交,如果相交的话,设交点坐标为p0(x0,y0).向量(p0p1)和(p0p2)的叉积为0,有(x1-x0)*(y2-y0)-(y1-y0)*(x2-x0)=0;同理,求出p0和p3p4直线的式子.…

两条直线可能有三种关系:1.共线     2.平行(不包括共线)    3.相交. 那给定两条直线怎么判断他们的位置关系呢.还是用到向量的叉积 例题:POJ 1269 题意:这道题是给定四个点p1, p2, p3, p4,直线L1,L2分别穿过前两个和后两个点.来判断直线L1和L2的关系 这三种关系一个一个来看: 1. 共线. 如果两条直线共线的话,那么另外一条直线上的点一定在这一条直线上.所以p3在p1p2上,所以用get_direction(p1, p2, p3)来判断p3相对于p1p2的关…