p308.ans1
======================
1.5707963267

=================
p308.cpp
======================
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

const double UMIN = -1.5, UMAX = 1.5, DU = 0.018;
const double PHIMIN = -15.0, PHIMAX = 15.0, DPHI = 0.028;

vector<float> a, b;

void read() {
    a.resize(3);
    b.resize(3);
    cin >> a >> b;
}

void solve() {
    // The key in this problem is that the tolerance is very large (+-0.1),
    // so we do NOT need a perfectly exact formula. Any method whose error
    // is safely below 0.05 will be accepted. Analytical solutions exist but
    // they are quite tricky to get right near the neck and when points are on
    // opposite horns. Therefore we choose a clean algorithmic approach.
    //
    // Imagine the hyperboloid as an infinite hourglass made of two funnels
    // glued together at their narrowest point. The narrow neck is exactly
    // the unit circle at z = 0. If you try to walk on this surface
    // using ordinary (x, y, z) coordinates the distances feel messy because
    // the circles get wider as you move away from the neck. So we change to
    // much more natural coordinates. We keep the usual angle phi = atan2(y, x)
    // which simply tells us how far around the current circle we are. For the
    // "height" we do not use z. Instead we use u = arcsinh(z). This u is like a
    // stretched height that automatically matches how the surface flares out.
    //
    // In these coordinates every point on the surface is given by:
    //
    //     x = cosh(u) * cos(phi),
    //     y = cosh(u) * sin(phi),
    //     z = sinh(u).
    //
    // At any fixed u the horizontal slice is a perfect circle whose radius
    // is exactly cosh(u). So moving around the circle (changing phi) costs
    // a distance of cosh(u) * dphi. Moving up or down (changing u)
    // costs a different amount that depends on how slanted the surface is
    // at that height. When we combine both directions the true distance element
    // on the surface simplifies to:
    //
    //     ds = sqrt(cosh(2 * u) * du * du + cosh(u) * cosh(u) * dphi * dphi).
    //
    // This single formula is all we need to measure any path correctly. Now we
    // build a fine rectangular grid in the flat (u, phi) plane. We should play
    // with the constants, but if u goes from -1.5 to +1.5 with step 0.018 and
    // phi goes from -15 to +15 with step 0.028, we get something that passes.
    // Each grid point corresponds to a real point on the hyperboloid. One
    // detail is that because phi repeats every 2 * pi we also consider several
    // copies of point B shifted by 2 * pi * k for k from -5 to +5. Between any
    // two neighbouring grid points we add an edge whose length is computed with
    // the ds formula above, using the midpoint of the segment for u. Dijkstra
    // then finds the shortest path on this graph. The result is a polygonal
    // line on the surface that follows the true geodesic closely.

    double xa = a[0], ya = a[1], za = a[2];
    double xb = b[0], yb = b[1], zb = b[2];

    double ua = asinh(za);
    double ub = asinh(zb);
    double dphi = atan2(yb, xb) - atan2(ya, xa);

    const double PI = acos(-1.0);
    dphi = fmod(dphi + 3 * PI, 2 * PI) - PI;

    int NU = (int)((UMAX - UMIN) / DU) + 15;
    int NP = (int)((PHIMAX - PHIMIN) / DPHI) + 15;

    vector<vector<double>> dist(NU, vector<double>(NP, 1e18));

    auto getiu = [&](double u) {
        return max(0, min(NU - 1, (int)round((u - UMIN) / DU)));
    };
    auto getip = [&](double p) {
        return max(0, min(NP - 1, (int)round((p - PHIMIN) / DPHI)));
    };

    int su = getiu(ua);
    int sp = getip(0.0);

    using T = tuple<double, int, int>;
    priority_queue<T, vector<T>, greater<T>> pq;
    pq.push({0.0, su, sp});
    dist[su][sp] = 0.0;

    int dirs[8][2] = {{-1, -1}, {-1, 0}, {-1, 1}, {0, -1},
                      {0, 1},   {1, -1}, {1, 0},  {1, 1}};

    while(!pq.empty()) {
        auto [cost, i, j] = pq.top();
        pq.pop();
        if(cost > dist[i][j]) {
            continue;
        }

        double u0 = UMIN + i * DU;
        double p0 = PHIMIN + j * DPHI;

        for(auto& d: dirs) {
            int ni = i + d[0];
            int nj = j + d[1];
            if(ni < 0 || ni >= NU || nj < 0 || nj >= NP) {
                continue;
            }

            double u1 = UMIN + ni * DU;
            double p1 = PHIMIN + nj * DPHI;
            double um = 0.5 * (u0 + u1);
            double du = u1 - u0;
            double dp = p1 - p0;
            double ds =
                sqrt(cosh(2 * um) * du * du + cosh(um) * cosh(um) * dp * dp);

            double nc = cost + ds;
            if(nc < dist[ni][nj]) {
                dist[ni][nj] = nc;
                pq.push({nc, ni, nj});
            }
        }
    }

    double ans = 1e18;
    int tu = getiu(ub);
    for(int k = -5; k <= 5; ++k) {
        double tp = dphi + 2 * PI * k;
        int basep = getip(tp);
        for(int di = -3; di <= 3; ++di) {
            for(int dj = -3; dj <= 3; ++dj) {
                int ti = tu + di;
                int tj = basep + dj;
                if(ti >= 0 && ti < NU && tj >= 0 && tj < NP) {
                    ans = min(ans, dist[ti][tj]);
                }
            }
        }
    }

    if(ans > 1e9) {
        ans = 0.0;
    }
    cout << setprecision(10) << fixed << ans << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}

=================
p308.in1
======================
1 0 0
0 1 0

=================
statement.txt
======================
308. Hyperboloid Distance
Time limit per test: 0.5 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



You are given two points A and B on the surface of the hyperboloid x2+y2-z2=1. Find the shortest distance between them along the surface.

Input
The input file contains six real numbers: xA, yA, zA, xB, yB and zB (coordinates of A and B), separated by spaces and/or line breaks (-1 ≤ zA, zB ≤ 1).

Output
Output the shortest distance between A and B along the surface of the hyperboloid. Your answer should be accurate within +-0.1.

Example(s)
sample input
sample output
1 0 0
0 1 0
1.5707963267

=================
