直线求交点两种方法

方法一：代数法。

对于l1,l2求出方程ax+by+c=0

解方程求交点。

方法二：几何法。

利用面积比值推出线段比值

再用相似三角形，进行向量放缩就可以了。

#include <iostream>
#include <cmath>
using namespace std;
const double eps=1e-8;
struct Tpoint
{
    double x,y;
    Tpoint(){}
    Tpoint(double _x,double _y){x=_x,y=_y;}
    void print()
    {
        printf("%.3lf %.3lf\n",x,y);
    }
};
struct Tline
{
    Tpoint p1,p2;
    double a,b,c;
    Tline(){}
    Tline(double x1,double y1,double x2,double y2){p1=Tpoint(x1,y1),p2=Tpoint(x2,y2);}
    void init()
    {
        a=p1.y-p2.y;
        b=p2.x-p1.x;
        c=p1.x*p2.y-p2.x*p1.y;
    }
};

int sig(double x)
{
    if(fabs(x)<=eps)return 0;
    if(x>eps)return 1;else return -1;
}

Tpoint operator -(Tpoint p1,Tpoint p2)
{
     return Tpoint(p1.x-p2.x,p1.y-p2.y);
}

Tpoint operator *(Tpoint p,double a)
{
    p.x*=a,p.y*=a;
    return p;
}

double operator *(Tpoint p1,Tpoint p2)
{
    return p1.x*p2.y-p2.x*p1.y;
}

Tpoint operator +(Tpoint p1,Tpoint p2)
{
    return Tpoint(p1.x+p2.x,p1.y+p2.y);
}

Tpoint lineintersect1(Tline l1,Tline l2)
{
    double ta=(l1.p2-l2.p1)*(l1.p1-l2.p1);
    double tb=(l2.p2-l1.p2)*(l1.p1-l1.p2);
    Tpoint tp;
    tp=(l2.p2-l2.p1)*(ta/(ta+tb))+l2.p1;
    return tp;
}

Tpoint lineintersect2(Tline l1,Tline l2)
{
    Tpoint p;
    double tp=l1.b*l2.a-l1.a*l2.b;
    if(sig(tp)==0)return p;
    p.x=(-l1.c*l2.b+l1.b*l2.c)/(-tp);
    p.y=(-l1.c*l2.a+l1.a*l2.c)/tp;
    return p;
}
int main()
{
    freopen("tmp.in","r",stdin);
    freopen("tmp.out","w",stdout);
    Tline l1,l2;
    double x1,x2,y1,y2;
    scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
    l1=Tline(x1,y1,x2,y2);l1.init();
    scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);
    l2=Tline(x1,y1,x2,y2);l2.init();
    Tpoint p1=lineintersect1(l1,l2);
    Tpoint p2=lineintersect2(l1,l2);
    p1.print(),p2.print();
    return 0;
}