学步园

如图所示，必然在AB，CD上存在两点e,f使A到D的时间最小，由此可以假设f点已知，由此确定下e点，所以采用嵌套三分

code

/*
ID: yueqiq
PROG: numtri
LANG: C++
*/
#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define lson l,m,rt << 1
#define rson m+1,r,rt<<1|1
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("input.txt","r",stdin)
#define writef freopen("output.txt","w",stdout)
const int maxn = 1001;
const int INF = 1111;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
const double pi = acos(-1.0);
const double eps= 1e-5;
using namespace std;
struct point{
    double x,y;
}A,B,C,D;
double P,Q,R;
double dist(point a,point b){
     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
//此时假设CD段上的点f为已知的,此时来枚举e点
double cal_ab(point a,point b,point f){
    point l=a,r=b,m,mid;
    double ans1,ans2;
    do{
        m.x=(l.x+r.x)/2.0;
        m.y=(l.y+r.y)/2.0;
        mid.x=(l.x+m.x)/2.0;
        mid.y=(l.y+m.y)/2.0;
        ans1=dist(a,m)/P+dist(m,f)/R;
        ans2=dist(a,mid)/P+dist(mid,f)/R;
        if(ans1<ans2) l=mid;
        else r=m;
    }while(fabs(ans1-ans2)>eps);
    return ans1;
}
double cal_abcd(point a,point b,point c,point d){
    point l=c,r=d,m,mid;
    double ans1,ans2;
    do{
        m.x=(l.x+r.x)/2.0;
        m.y=(l.y+r.y)/2.0;
        mid.x=(l.x+m.x)/2.0;
        mid.y=(l.y+m.y)/2.0;
        ans1=dist(m,d)/Q+cal_ab(a,b,m);
        ans2=dist(mid,d)/Q+cal_ab(a,b,mid);
        if(ans1<ans2) l=mid;
        else r=m;
    }while(fabs(ans1-ans2)>eps);
    return ans1;
}
int main()
{
    int cas;
    SD(cas);
    while(cas--){
        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\n",cal_abcd(A,B,C,D));
    }
    return 0;
}