三分

雨巨点名的三分题……本菜鸡还是不太懂，观摩其他大佬的代码终于有点理解。
我们先三分出从A点到AB中的某个点X，作为出发点，然后，再三分出从X到CD的某个点Y，再从Y直接到D，这样，我们就可以求出最小的值了。
路径的话，存在4条路，分别代表走不走传送带。
不走AB带就直接走到B，不走CD带就直接从x走到D，直接码三分的板子就行了，向量求起来有点复杂，要仔细。
精度问题，建议double的直接循环100或者1000次，看题目规模去改，别搞超时了。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define all(vv) (vv).begin(), (vv).end()
#define endl "\n"
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll tmp = x > 0 ? x : -x; if (x < 0)putchar('-');    int cnt = 0;    while (tmp > 0) { F[cnt++] = tmp % 10 + '0';        tmp /= 10; }    while (cnt > 0)putchar(F[--cnt]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const double eps = 1e-4;
double p, q, r;
struct node {
    double x, y;
}a, b, c, d, e, f;

double dis(node a, node b) {
    return sqrt(eps + (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
double check1(double x) {
    f = { c.x + (x / dis(c,d) * (d.x - c.x)),c.y + (x / dis(c,d) * (d.y - c.y)) };
    return dis(e, f) / r + (dis(c, d) - x) / q;
}
double check(double x) {
    e = { a.x + x / dis(a,b) * (b.x - a.x),a.y + x / dis(a,b) * (b.y - a.y) };
    double l = 0, r = dis(c, d);
    for (int i = 1; i <= 1000; i++) {
        double lm = l + (r - l) / 3, rm = r - (r - l) / 3;
        if (check1(lm) >= check1(rm))l = lm;
        else r = rm;
    }
    return check1(l) + x / p;
}
int main() {
    cin >> a.x >> a.y >> b.x >> b.y >> c.x >> c.y >> d.x >> d.y;
    cin >> p >> q >> r;
    double l = 0, r = dis(a, b);
    for (int i = 1; i <= 1000; i++) {
        double lm = l + (r - l) / 3, rm = r - (r - l) / 3;
        if (check(lm) >= check(rm))l = lm;
        else r = rm;
    }
    printf("%.2f", check(l));
    return 0;
}