hdu3400 两重三分

题意:
     题意给你两个公路 A-B C-D 和三个速度V(ab) V(cd) 和 V(两条公路之间) 问你从A到D的最短时间是多少.


思路:
   一开始暴力了其中的一条边,每次加0.01,另一条边用的三分,结果wa掉了,感觉不wa暴力一条边时间上也够呛,后来看了下题解,人家用的是两重三分,就是三分其中一条边,当对于最外层的那个三分的某两个点也就是 mid mmid,我们在三分两次,取得最优,
确实如此,因为后来想了想,对于整体来说,总函数里面有两个未知数,无法确定是他的性质,

但是如果我们分开来想,分成两部分,那么他们就含有凸性(或凹性)了,这样我们就可以三分在短时间内找到精度满足条件的解..

#include<stdio.h>
#include<math.h>

#define eps 0.0001

typedef struct
{
   double x ,y;
}NODE;

NODE A ,B ,C ,D;
double P ,Q ,R;

double dis(NODE X ,NODE Y)
{
   double tmp = pow(X.x - Y.x ,2.0) + pow(X.y - Y.y ,2.0);
   return sqrt(tmp);
}

double CD_3F(NODE now)
{
   NODE low ,up ,mid ,mmid;
   double t1 ,t2;
   low = C ,up = D;
   while(1)
   {
      mid.x = (low.x + up.x) / 2;
      mid.y = (low.y + up.y) / 2;
      t1 = dis(now ,mid) / R + dis(mid ,D) / Q;
      
      mmid.x = (mid.x + up.x) / 2;
      mmid.y = (mid.y + up.y) / 2;
      t2 = dis(now ,mmid) / R + dis(mmid ,D) / Q;
      
      if(t1 > t2) low = mid;
      else up = mmid;
      
      if(dis(low ,up) < eps) break;
   }
   return t2;
}

double AB_3F()
{
   NODE low ,up ,mid ,mmid;
   low = A ,up = B;
   double t1 ,t2;
   while(1)
   {       //puts("ok");
      mid.x = (low.x + up.x) / 2;
      mid.y = (low.y + up.y) / 2;
      t1 = dis(A ,mid) / P + CD_3F(mid);
      
      mmid.x = (mid.x + up.x) / 2;
      mmid.y = (mid.y + up.y) / 2;
      t2 = dis(A ,mmid) / P + CD_3F(mmid);
      
      if(t1 > t2) low = mid;
      else up = mmid;
      
      if(dis(low ,up) < eps) break;
   }
   return t1;
}

int main ()
{
   int t;
   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);
      printf("%.2lf
" ,AB_3F());
   }
   return 0;
}