poj 1269 线段相交/平行

模板题

注意原题中说的线段其实要当成没有端点的直线。被坑了= =

    #include <cmath>
    #include <cstdio>
    #include <iostream>
    #include <cstring>
    using namespace std;

    #define eps 1e-8
    #define PI acos(-1.0)//3.14159265358979323846
    //判断一个数是否为0,是则返回true,否则返回false
    #define zero(x)(((x)>0?(x):-(x))<eps)
    //返回一个数的符号，正数返回1，负数返回2，否则返回0
    #define _sign(x)((x)>eps?1:((x)<-eps?2:0))

   struct point
   {
       double x,y;
       point(){}
       point(double xx,double yy):x(xx),y(yy)
       {}
   };
   struct line
   {
       point a,b;
       line(){}      //默认构造函数
       line(point ax,point bx):a(ax),b(bx)
       {}
   };//直线通过的两个点，而不是一般式的三个系数

     int n;
     double ax1,ay1,ax2,ay2,bx1,by1,bx2,by2;

    //求矢量[p0,p1],[p0,p2]的叉积
    //p0是顶点
    //若结果等于0，则这三点共线
    //若结果大于0，则p0p2在p0p1的逆时针方向
    //若结果小于0，则p0p2在p0p1的顺时针方向
    double xmult(point p1,point p2,point p0)
    {
        return(p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
    }
    //计算dotproduct(P1-P0).(P2-P0)
    double dmult(point p1,point p2,point p0)
    {
        return(p1.x-p0.x)*(p2.x-p0.x)+(p1.y-p0.y)*(p2.y-p0.y);
    }
    //两点距离
    double distance(point p1,point p2)
    {
        return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
    }
    //判三点共线
    int dots_inline(point p1,point p2,point p3)
    {
        return zero(xmult(p1,p2,p3));
    }
    //判点是否在线段上,包括端点
    int dot_online_in(point p,line l)
    {
        return zero(xmult(p,l.a,l.b))&&(l.a.x-p.x)*(l.b.x-p.x)<eps&&(l.a.y-p.y)*(l.b.y-p.y)<eps;
    }
    //判点是否在线段上,不包括端点
    int dot_online_ex(point p,line l)
    {
        return dot_online_in(p,l)&&(!zero(p.x-l.a.x)||!zero(p.y-l.a.y))&&(!zero(p.x-l.b.x)||!zero(p.y-l.b.y));
    }
    //判两点在线段同侧,点在线段上返回0
    int same_side(point p1,point p2,line l)
    {
        return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)>eps;
    }
    //判两点在线段异侧,点在线段上返回0
    int opposite_side(point p1,point p2,line l)
    {
        return xmult(l.a,p1,l.b)*xmult(l.a,p2,l.b)<-eps;
    }
    //判两直线平行
    int parallel(line u,line v)
    {
        return zero((u.a.x-u.b.x)*(v.a.y-v.b.y)-(v.a.x-v.b.x)*(u.a.y-u.b.y));
    }
    //判两直线垂直
    int perpendicular(line u,line v)
    {
        return zero((u.a.x-u.b.x)*(v.a.x-v.b.x)+(u.a.y-u.b.y)*(v.a.y-v.b.y));
    }
    //判两线段相交,包括端点和部分重合
    int intersect_in(line u,line v)
    {
        if(!dots_inline(u.a,u.b,v.a)||!dots_inline(u.a,u.b,v.b))
            return!same_side(u.a,u.b,v)&&!same_side(v.a,v.b,u);
        return dot_online_in(u.a,v)||dot_online_in(u.b,v)||dot_online_in(v.a,u)||dot_online_in(v.b,u);
    }
    //判两线段相交,不包括端点和部分重合
    int intersect_ex(line u,line v)
    {
        return opposite_side(u.a,u.b,v)&&opposite_side(v.a,v.b,u);
    }
    //计算两直线交点,注意事先判断直线是否平行!
    //线段交点请另外判线段相交(同时还是要判断是否平行!)
    point intersection(line u,line v)
    {
        point ret=u.a;
        double t=((u.a.x-v.a.x)*(v.a.y-v.b.y)-(u.a.y-v.a.y)*(v.a.x-v.b.x))/((u.a.x-u.b.x)*(v.a.y-v.b.y)-(u.a.y-u.b.y)*(v.a.x-v.b.x));
        ret.x+=(u.b.x-u.a.x)*t;
        ret.y+=(u.b.y-u.a.y)*t;
        return ret;
    }

   int main()
   {
       cin>>n;
       cout<<"INTERSECTING LINES OUTPUT"<<endl;
       for (int i=;i<=n;i++)
       {
         cin>>ax1>>ay1>>ax2>>ay2>>bx1>>by1>>bx2>>by2;
         point a1(ax1,ay1);  point a2(ax2,ay2);
         point b1(bx1,by1);  point b2(bx2,by2);
         line l1=line(point(ax1,ay1),point(ax2,ay2));
         line l2=line(point(bx1,by1),point(bx2,by2));
         if ((dots_inline(a1,a2,b1)>)&&(dots_inline(a1,a2,b2)>))
             cout<<"LINE";
         else if (parallel(l1,l2)>)     cout<<"NONE";
         else
         {
             point tm=intersection(l1,l2);
             cout<<"POINT ";
             printf("%.2f %.2f",tm.x,tm.y);
         }
         cout<<endl;
       }
       cout<<"END OF OUTPUT"<<endl;
       return ;
   }