poj 1269 知识点：直线相交判断，求相交交点

#include<iostream>
using namespace std;
const double epx=1e-10;
struct Point
{
    double x;
    double y;
};
//求解二元一次方程
Point solve(double a1,double b1,double c1,double a2,double b2,double c2)
{
    Point p;
    p.x=(c1*b2-c2*b1)/(a2*b1-a1*b2);
    p.y=(a2*c1-a1*c2)/(a1*b2-a2*b1);
    return p;
}
//p1p3,p1p2的叉积
double direction(Point p1,Point p2,Point p3)
{
    return (p3.x-p1.x)*(p2.y-p1.y)-(p2.x-p1.x)*(p3.y-p1.y);
}
//判断两直线的关系
/*
在这里有三种关系：1共线 2平行 3相交
1 共线可通过叉积来判断
2 平行通过向量来判断
3 通了上面2种情况的其他情况
  求交点可通过叉积及解二元一次方程来求解
*/
int N;
Point p1,p2,p3,p4;
Point p0;//交点
double a1,b1,c1,a2,b2,c2;
int main()
{
    scanf("%d",&N);
    int i;
    printf("INTERSECTING LINES OUTPUT\n");
    for(i=0;i<N;++i)
    {
        scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &p1.x, &p1.y, &p2.x, &p2.y, &p3.x, &p3.y, &p4.x, &p4.y);
        
        if(direction(p3,p4,p1)==0 && direction(p3,p4,p2)==0)//共线
            printf("LINE\n");
        else
        {
            if( ((p1.x-p2.x)*(p3.y-p4.y)-(p1.y-p2.y)*(p3.x-p4.x))==0 )//平行
                printf("NONE\n");
            else
            {
                a1=p1.y-p2.y;b1=p2.x-p1.x;c1=p1.x*p2.y-p2.x*p1.y;
                a2=p3.y-p4.y;b2=p4.x-p3.x;c2=p3.x*p4.y-p4.x*p3.y;
                p0=solve(a1,b1,c1,a2,b2,c2);
                printf("POINT %.2f %.2f\n",p0.x,p0.y);
            }
        }
    }
    printf("END OF OUTPUT\n");
    return 0;
}