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;}