hdu 3400 Line belt

题意：给你两条线段AB，CD；然后给你在AB，CD上的速度P，Q，在其它部分的速度是R，然后求A到D的最短时间。

思路：用三分枚举从AB线段上离开的点，然后再用三分枚举在CD的上的点找到最优点，求距离和时间就可以。

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 using namespace std;
 const double eps=1e-;

 int t;
 double p,q,r;
 struct point
 {
      double x,y;
 }a,b,c,d;

 double sqr(double x)
 {
     return x*x;
 }

 double dis(double x1,double y1,double x2,double y2)
 {
     return sqrt(sqr(x1-x2)+sqr(y1-y2)+eps);
 }

 double ok(double t1)
 {
     point pos;
     pos.x=a.x+(b.x-a.x)*((t1*p)/dis(a.x,a.y,b.x,b.y));
     pos.y=a.y+(b.y-a.y)*((t1*p)/dis(a.x,a.y,b.x,b.y));
     double l1=,r1=dis(c.x,c.y,d.x,d.y)/q;
     while(r1-l1>eps)
     {
         double mid1=(r1+l1)/;
         double mid2=(mid1+r1)/;
         double x1=d.x+(c.x-d.x)*((mid1*q)/dis(c.x,c.y,d.x,d.y));
         double y1=d.y+(c.y-d.y)*((mid1*q)/dis(c.x,c.y,d.x,d.y));
         double x2=d.x+(c.x-d.x)*((mid2*q)/dis(c.x,c.y,d.x,d.y));
         double y2=d.y+(c.y-d.y)*((mid2*q)/dis(c.x,c.y,d.x,d.y));
         double d1=dis(pos.x,pos.y,x1,y1)/r;
         double d2=dis(pos.x,pos.y,x2,y2)/r;
         if(d1+mid1<=d2+mid2)
         {
             r1=mid2;
         }
         else
             l1=mid1;
     }
     double xx=d.x+(c.x-d.x)*(l1*q)/dis(c.x,c.y,d.x,d.y);
     double yy=d.y+(c.y-d.y)*(l1*q)/dis(c.x,c.y,d.x,d.y);
     return dis(pos.x,pos.y,xx,yy)/r+t1+l1;
 }

 int main()
 {
     scanf("%d",&t);
     while(t--)
     {
         scanf("%lf%lf%lf%lf",&a.x,&a.y,&b.x,&b.y);
         scanf("%lf%lf%lf%lf",&c.x,&c.y,&d.x,&d.y);
         scanf("%lf%lf%lf",&p,&q,&r);
         double ll=,rr=dis(a.x,a.y,b.x,b.y)/p;
         while(rr-ll>eps)
         {
             double mid1=(ll+rr)/;
             double mid2=(mid1+ll)/;
             if(ok(mid2)>=ok(mid1))
             {
                 ll=mid2;
             }
             else
                 rr=mid1;
         }
         printf("%.2lf\n",ok(ll));
     }
 }