求直线(线段)与直线(线段)的交点

知识储备：
叉乘：http://blog.csdn.net/nightmare_ak/article/details/77199940
定比分点法：http://blog.csdn.net/nightmare_ak/article/details/77917293

对于线段，只要先判断是否相交，就可以转化成直线求交点了

附模板：

#include<cstdio>#include<algorithm>#include<cmath>using namespace std;const int MAXN = 20 + 5;struct Pos{    double _x, _y;    Pos(double x=0.0,double y=0.0):_x(x),_y(y){}}up[MAXN], down[MAXN];double Cross(Pos p1, Pos p2, Pos p0)//叉乘{    return (p1._x - p0._x)*(p2._y - p0._y) - (p2._x - p0._x)*(p1._y - p0._y);}bool isColide(Pos p1,Pos p2,Pos p3,Pos p4)//判断是否相交,直线p1p2与线段p3p4{    double tmp = Cross(p2, p3, p1)*Cross(p2, p4, p1);    return tmp < 0.0 || fabs(tmp) < 1e-6;}Pos getPoint(Pos p1, Pos p2, Pos p3, Pos p4)//求出交点,直线(线段)p1p2与直线(线段)p3p4{//t=lamta/(1+lamta)    double x1 = p1._x, y1 = p1._y;    double x2 = p2._x, y2 = p2._y;    double x3 = p3._x, y3 = p3._y;    double x4 = p4._x, y4 = p4._y;    double t = ((x2 - x1)*(y3 - y1) - (x3 - x1)*(y2 - y1)) / ((x2 - x1)*(y3 - y4) - (x3 - x4)*(y2 - y1));    return Pos(x3 + t*(x4 - x3), y3 + t*(y4 - y3));}int main(){    return 0;}