E

题意略。

本题先枚举是先去第一条线还是第二条线，再三分套三分也能过。

但究竟为啥能呢？我并没有太看出来这个问题具有什么凸性质 ...

#include <bits/stdc++.h>
#define INF 2000000000
#define MOD 1000000007
#define MAXN 200005
#define REP(temp, init_val, end_val) for (int temp = init_val; temp <= end_val; ++temp)
#define REPR(temp, init_val, end_val) for (int temp = init_val; temp >= end_val; --temp)

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> intpair;
int read(){
    int f = 1, x = 0;
    char c = getchar();
    while (c < '0' || c > '9'){if(c == '-') f = -f; c = getchar();}
    while (c >= '0' && c <= '9')x = x * 10 + c - '0', c = getchar();
    return f * x; 
}
double a, b, c, d, xx0, xx1, yy0, yy1;
void init(){
    a = read(), b = read(), c = read(), d = read();
    xx0 = read(), yy0 = read(), xx1 = read(), yy1 = read();
}
inline double distot1(double sf1, double sf2){
    return hypot(xx0 - a * sf1, yy0 - b * sf1) + hypot(a * sf1 - c * sf2, b * sf1 - d * sf2) + 
        hypot(xx1 - c * sf2, yy1 - d * sf2);
}
inline double distot2(double sf1, double sf2){
    return hypot(xx1 - a * sf1, yy1 - b * sf1) + hypot(a * sf1 - c * sf2, b * sf1 - d * sf2) + 
        hypot(xx0 - c * sf2, yy0 - d * sf2);
}
double calc1(double sf1){
    double lb2 = 0.0, rb2 = 1e3;
    REP(T, 1, 50){
        double len = (rb2 - lb2) / 3;
        double mid = lb2 + len;
        double mid2 = rb2 - len;
        if (distot1(sf1, mid) < distot1(sf1, mid2))
            rb2 = mid2;
        else lb2 = mid;
    }
    return distot1(sf1, lb2);
}
double calc2(double sf1){
    double lb2 = 0.0, rb2 = 1e3;
    REP(T, 1, 50){
        double len = (rb2 - lb2) / 3;
        double mid = lb2 + len;
        double mid2 = rb2 - len;
        if (distot2(sf1, mid) < distot2(sf1, mid2))
            rb2 = mid2;
        else lb2 = mid;
    }
    return distot2(sf1, lb2);
}
void solve(){
    double dis1 = hypot(a, b);
    double dis2 = hypot(c, d);
    a /= dis1, b /= dis1;
    c /= dis2, d /= dis2;

    double ans = 1e7;
    double lb1 = 0.0, rb1 = 1e3;

    REP(T1, 1, 50){
        double len = (rb1 - lb1) / 3;
        double mid = lb1 + len;
        double mid2 = rb1 - len;
        if (calc1(mid) < calc1(mid2))
            rb1 = mid2;
        else lb1 = mid;
        // cout << lb1 << " " << rb1 << endl;
    }

    ans = min(ans, calc1(lb1));

    lb1 = 0.0, rb1 = 1e3;

    REP(T1, 1, 50){
        double len = (rb1 - lb1) / 3;
        double mid = lb1 + len;
        double mid2 = rb1 - len;
        if (calc2(mid) < calc2(mid2))
            rb1 = mid2;
        else lb1 = mid;
    }

    ans = min(ans, calc2(lb1));

    printf("%.3lf\n", ans);
}
int main(){
    int T = 1;
    while (T--){
        init();
        solve();
    }
    return 0;
}