思路:
首先在AB上三分一個點出來,然後再在CD上三分一個點讓值最小兩次三分重疊起來。結果應加上一個精度。
#include<stdio.h>#include<math.h>
#define eps 1e-9
struct point
{
double x;
double y;
};
point A,B,C,D,M1,M2;
double P,Q,R;
double dis(point a,point b)
{
return sqrt(eps+(a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double cal2(double len)
{
double d1,d2,k,t1,t2;
d1=len,d2=dis(C,D);
k=d1/d2;
M2.x=(C.x-D.x)*k+D.x;
M2.y=(C.y-D.y)*k+D.y;
t1=dis(M1,M2)/R;
t2=len/Q;
return (t1+t2);
}
double cal1(double len)
{
int i;
double d1,d2,k,t1,tx,ty;
d1=len,d2=dis(A,B);
k=d1/d2;
M1.x=(B.x-A.x)*k+A.x;
M1.y=(B.y-A.y)*k+A.y;
t1=len/P;
double left,right,mid1,mid2;
left=0,right=dis(C,D);
for(i=1;i<=100;i++)
{
mid1=(2*left+right)/3;
mid2=(left+2*right)/3;
tx=cal2(mid1);
ty=cal2(mid2);
if(tx>ty)
{
left=mid1;
}
else
{
right=mid2;
}
}
return t1+cal2(left);
}
void triple()
{
int i;
double mid1,mid2,left,right,t1,t2;
left=0,right=dis(A,B);
for(i=1;i<=100;i++)
{
mid1=(left*2+right)/3;
mid2=(left+2*right)/3;
t1=cal1(mid1);
t2=cal1(mid2);
if(t1>t2)
{
left=mid1;
}
else
{
right=mid2;
}
}
printf("%.2lf\n",cal1(left));
}
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);
triple();
}
return 0;
}