Intersecting Lines Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12421   Accepted: 5548 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…

用的是初中学的方法 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define eps 1e-8 using namespace std; struct Point { double x,y; Point() {}; Point(double xx,double yy) { x=xx; y=yy; }…

题目链接:POJ 1269 Problem 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 a line becau…

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…

题目链接 题意 : 给你两条线段的起点和终点,一共四个点,让你求交点坐标,如果这四个点是共线的,输出“LINE”,如果是平行的就输出“NONE”. 思路 : 照着ZN留下的模板果然好用,直接套上模板了事儿,不过在判断是否共线的时候,其实还有另一种方法,直接将平行和共线一起判断了,我是判断三个点三个点的判断是否是共线. #include <stdio.h> #include <string.h> #include <iostream> using namespace st…

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…

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.…

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…