题解 | 两直线交点

#include<bits/stdc++.h>
using namespace std;
struct point {
    double x,y;
    point(double A,double B){
        x = A, y = B;
    }
    point() = default;
};
struct line {
    point point_A,point_B;
    line(point A,point B){
        point_A = A,point_B = B;
    }
    line() = default;
};
point findMeetingPoint(line line_A,line line_B){
    point p;
    double xa = line_A.point_A.x,ya = line_A.point_A.y;
    double xb = line_A.point_B.x,yb = line_A.point_B.y;
    double xc = line_B.point_A.x,yc = line_B.point_A.y;
    double xd = line_B.point_B.x,yd = line_B.point_B.y;
  点P弄出来，点ABCD弄出来。
    if(fabs((yc-yd)*(xa-xb) - (xc-xd)*(ya-yb)) < 1e-10) {
        double distAtoCD = fabs((xd-xc)*(ya-yc) -(xa-xc)*(yd-yc));
        double distCtoAB = fabs((xc-xa)*(ya-yb)-(yc-ya)*(xa-xb));
        if(distAtoCD<1e-10&&distCtoAB<1e-10){
            return point(-1,-1);
        }
        else{
            return point(-1,-1);
        }
    }
     double x0 = ((xc - xd) * (xb * ya - xa * yb) - (xa - xb) *(xd * yc - xc * yd)) / (-(yc - yd) * (xa - xb) + (xc - xd) * (ya - yb)); // 解方程
    double y0 = ((xa * yb - ya * xb) * (yc - yd) - (ya - yb) *(xc * yd - yc * xd)) / ((yc - yd) * (xa - xb) - (xc - xd) * (ya - yb));
    p.x = x0, p.y = y0;
    return p;
}
int main(){
    point A,B,C,D;
    cin >> A.x >>A.y>>B.x>>B.y>>C.x>>C.y>>D.x>>D.y;
    line AB = line(A,B);
    line CD = line(C,D);
    point ans = findMeetingPoint(AB,CD);
    cout << fixed <<setprecision(6);
    cout << ans.x <<" " << ans.y;
    return 0;
}