poj 1269 Intersecting Lines

题目链接：http://poj.org/problem?id=1269

题目大意：给出四个点的坐标x1,y1,x2,y2,x3,y3,x4,y4,前两个形成一条直线，后两个坐标形成一条直线。然后问你是否平行，重叠或者相交，如果相交，求出交点坐标。

算法：二维几何直线相交+叉积

解法：先用叉积判断是否相交，如果相交的话，设交点坐标为p0(x0,y0)。向量（p0p1）和(p0p2)的叉积为0，有(x1-x0)*(y2-y0)-(y1-y0)*(x2-x0)=0;同理，求出p0和p3p4直线的式子。然后联立求解x0,y0。平行或重叠的情况就自己YY了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#define inf 0x7fffffff
#define exp 1e-10
#define PI 3.141592654
using namespace std;
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y() {}
};
typedef Point Vector;
double cross(Vector A,Vector B)
{
    return A.x*B.y-A.y*B.x;
}
Point GetLineIntersection(Point P,Vector v,Point Q,Vector w)
{
    Point uu;
    Vector u=Point(P.x-Q.x , P.y-Q.y);
    double t=cross(w,u)/cross(v,w);
    uu.x=P.x+v.x*t;
    uu.y=P.y+v.y*t;
    return uu;
}//调用训练指南上这个函数怎么错了，我写错了吗
int main()
{
    int n;
    double x1,y1,x2,y2,x3,y3,x4,y4;
    //cin>>n;
    while (cin>>n)
    {
        printf("INTERSECTING LINES OUTPUT\n");
        while (n--) {
        cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;
        Point P,Q;
        P.x=x1 ;P.y=y1 ;
        Q.x=x3 ;Q.y=y3 ;
        Vector v,w;
        v.x=x2-x1 ;v.y=y2-y1;
        w.x=x4-x3 ;w.y=y4-y3;
        if (cross(v,w)!=0)
        {
            //Vector vv=GetLineIntersection(P,v,Q,w);
            double a1,b1,c1;
            double a2,b2,c2;
            a1=y1-y2 ;b1=x2-x1 ;c1=x1*y2-y1*x2;
            a2=y3-y4 ;b2=x4-x3 ;c2=x3*y4-y3*x4;
            double x0=(b1*c2-b2*c1)/(b2*a1-b1*a2);
            double y0=(a2*c1-a1*c2)/(a1*b2-a2*b1);
            printf("POINT %.2f %.2f\n",x0,y0);
        }
        else
        {
            if (fabs(v.x)<=exp && fabs(w.x)<=exp)
            {
                if (fabs(x1-x3)<=exp)
                printf("LINE\n");
                else printf("NONE\n");
            }
            else if (fabs(v.y)<=exp && fabs(w.y)<=exp)
            {
                if (fabs(y1-y3)<=exp)
                printf("LINE\n");
                else printf("NONE\n");
            }
            else
            {
                if (fabs((y3-w.y/w.x*x3)-(y1-w.y/w.x*x1))<=exp)
                printf("LINE\n");
                else printf("NONE\n");
            }
        }
        }
        printf("END OF OUTPUT\n");
    }
    return 0;
}