## 1) Abridged problem statement

Given two points \(A(x_A,y_A,z_A)\) and \(B(x_B,y_B,z_B)\) on the hyperboloid
\[
x^2+y^2-z^2=1,
\]
find the length of the shortest path **along the surface** (a geodesic) connecting them.  
Input: six real numbers (coordinates), with \(-1 \le z_A,z_B \le 1\).  
Output: the surface distance, with error tolerance \(\pm 0.1\).

---

## 2) Detailed editorial (explaining the provided approach)

### Key idea: avoid exact geodesic formulas; use a numerical shortest-path
The exact analytic geodesic on a hyperboloid can be derived but is easy to implement incorrectly near the “neck” and when the points lie on opposite sides. Fortunately, the required accuracy is only \(\pm 0.1\), so we can approximate the surface by a fine grid and run Dijkstra’s algorithm to obtain a near-geodesic distance.

### Step 1: Parametrize the hyperboloid with natural coordinates \((u,\phi)\)
Use:
- \(\phi = \mathrm{atan2}(y,x)\) (angle around the axis),
- \(u = \mathrm{asinh}(z)\) (a “stretched” vertical coordinate).

With these, every surface point is:
\[
x=\cosh(u)\cos\phi,\quad y=\cosh(u)\sin\phi,\quad z=\sinh(u).
\]
Indeed:
\[
x^2+y^2-z^2=\cosh^2(u)-\sinh^2(u)=1.
\]

So the hyperboloid becomes a “rectangle-like” domain in \((u,\phi)\) space.

### Step 2: Surface metric (distance element)
Compute the first fundamental form (or accept the known result for this parametrization). The infinitesimal distance on the surface is:
\[
ds^2 = \cosh(2u)\,du^2 + \cosh^2(u)\,d\phi^2,
\]
so
\[
ds = \sqrt{\cosh(2u)\,du^2 + \cosh^2(u)\,d\phi^2}.
\]

This formula lets us assign a length to a small move in \((u,\phi)\).

### Step 3: Discretize \((u,\phi)\) into a grid graph
Choose ranges and steps (constants in the code):

- \(u \in [U_{\min},U_{\max}] = [-1.5, 1.5]\) with step \(DU=0.018\)
- \(\phi \in [\Phi_{\min},\Phi_{\max}] = [-15, 15]\) with step \(DPHI=0.028\)

Each grid node \((i,j)\) corresponds to:
\[
u = U_{\min} + i\cdot DU,\quad \phi = \Phi_{\min} + j\cdot DPHI.
\]

Connect each node to its 8 neighbors (including diagonals). For an edge from \((u_0,\phi_0)\) to \((u_1,\phi_1)\), approximate its length by evaluating the metric at the midpoint \(u_m=(u_0+u_1)/2\):
\[
\Delta u = u_1-u_0,\quad \Delta\phi = \phi_1-\phi_0,
\]
\[
\text{edge length} \approx \sqrt{\cosh(2u_m)\Delta u^2 + \cosh^2(u_m)\Delta\phi^2}.
\]

This turns the problem into shortest path on a weighted graph.

### Step 4: Fix periodicity in \(\phi\)
Angle \(\phi\) is periodic modulo \(2\pi\). The code handles this by:
- Shifting coordinates so that point \(A\) is treated as \(\phi=0\),
- Using \(\Delta\phi\) between \(A\) and \(B\), normalized to \((-\pi,\pi]\),
- Considering multiple “copies” of \(B\) at \(\Delta\phi + 2\pi k\) for \(k \in [-5,5]\) because the grid’s \(\phi\) span is wide (\([-15,15]\)).

This ensures Dijkstra can find routes that wrap around the axis.

### Step 5: Run Dijkstra from \(A\), read best distance near \(B\)
- Start node: \(u=\mathrm{asinh}(z_A)\), \(\phi=0\).
- Dijkstra computes shortest distances to all grid nodes.
- Target: \(u=\mathrm{asinh}(z_B)\), \(\phi=\Delta\phi + 2\pi k\) for several \(k\).
- Because the true point may not land exactly on a grid node, the code checks a small \(7\times 7\) neighborhood around the nearest target cell and takes the minimum.

### Complexity
Let \(N = NU \cdot NP\) nodes. Each has up to 8 edges. Dijkstra is:
\[
O(N\log N).
\]
With the chosen grid sizes, this fits in time in C++ for the given limits, and the coarse tolerance makes it acceptable.

---

## 3) C++ Solution
```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;
}
```

---

## 4) Python solution (same method) with detailed comments

```python
import math
import heapq
import sys

# Same discretization constants as the C++ solution
UMIN, UMAX, DU = -1.5, 1.5, 0.018
PHIMIN, PHIMAX, DPHI = -15.0, 15.0, 0.028
PI = math.acos(-1.0)

def clamp(x, lo, hi):
    return lo if x < lo else hi if x > hi else x

def solve():
    data = list(map(float, sys.stdin.read().split()))
    xa, ya, za, xb, yb, zb = data

    # Convert to u = asinh(z)
    ua = math.asinh(za)
    ub = math.asinh(zb)

    # Principal angular difference between B and A
    dphi = math.atan2(yb, xb) - math.atan2(ya, xa)
    # Normalize to (-pi, pi]
    dphi = math.fmod(dphi + 3 * PI, 2 * PI) - PI

    # Grid sizes (+15 padding as in C++)
    NU = int((UMAX - UMIN) / DU) + 15
    NP = int((PHIMAX - PHIMIN) / DPHI) + 15

    # Map continuous u to nearest grid index, clamped
    def getiu(u):
        idx = int(round((u - UMIN) / DU))
        return clamp(idx, 0, NU - 1)

    # Map continuous phi to nearest grid index, clamped
    def getip(p):
        idx = int(round((p - PHIMIN) / DPHI))
        return clamp(idx, 0, NP - 1)

    # Dijkstra distance array (use a large number as infinity)
    INF = 1e100
    dist = [[INF] * NP for _ in range(NU)]

    # Start at (ua, phi=0)
    su = getiu(ua)
    sp = getip(0.0)
    dist[su][sp] = 0.0

    # Min-heap entries are (cost, i, j)
    pq = [(0.0, su, sp)]

    # 8-neighborhood
    dirs = [(-1, -1), (-1, 0), (-1, 1),
            (0, -1),          (0, 1),
            (1, -1),  (1, 0), (1, 1)]

    while pq:
        cost, i, j = heapq.heappop(pq)
        if cost != dist[i][j]:
            continue

        # Convert indices to coordinate values
        u0 = UMIN + i * DU
        p0 = PHIMIN + j * DPHI

        for di, dj in dirs:
            ni, nj = i + di, j + dj
            if not (0 <= ni < NU and 0 <= nj < NP):
                continue

            u1 = UMIN + ni * DU
            p1 = PHIMIN + nj * DPHI

            # Midpoint approximation for u in the metric
            um = 0.5 * (u0 + u1)

            du = u1 - u0
            dp = p1 - p0

            # ds = sqrt( cosh(2u) du^2 + cosh(u)^2 dphi^2 )
            ds = math.sqrt(math.cosh(2.0 * um) * du * du +
                           (math.cosh(um) ** 2) * dp * dp)

            nc = cost + ds
            if nc < dist[ni][nj]:
                dist[ni][nj] = nc
                heapq.heappush(pq, (nc, ni, nj))

    # Evaluate best distance near B, considering periodic copies in phi
    ans = INF
    tu = getiu(ub)

    for k in range(-5, 6):
        tp = dphi + 2 * PI * k
        basep = getip(tp)

        # 7x7 neighborhood around the nearest cell
        for di in range(-3, 4):
            for dj in range(-3, 4):
                ti, tj = tu + di, basep + dj
                if 0 <= ti < NU and 0 <= tj < NP:
                    ans = min(ans, dist[ti][tj])

    if ans > 1e50:
        ans = 0.0

    # Print with fixed precision
    print(f"{ans:.10f}")

if __name__ == "__main__":
    solve()
```

---

## 5) Compressed editorial

Parametrize the hyperboloid \(x^2+y^2-z^2=1\) by \(u=\mathrm{asinh}(z)\), \(\phi=\mathrm{atan2}(y,x)\):
\[
(x,y,z)=(\cosh u\cos\phi,\ \cosh u\sin\phi,\ \sinh u).
\]
Then the surface metric is
\[
ds^2=\cosh(2u)\,du^2+\cosh^2(u)\,d\phi^2.
\]
Discretize \((u,\phi)\) into a fine grid over \(u\in[-1.5,1.5]\), \(\phi\in[-15,15]\), connect each cell to its 8 neighbors, and weight edges by
\[
\Delta s \approx \sqrt{\cosh(2u_m)\Delta u^2+\cosh^2(u_m)\Delta\phi^2}
\]
(midpoint \(u_m\)). Run Dijkstra from \(A\) (set \(\phi_A=0\)) and take the minimum distance to \(B\) over multiple \(\phi\) copies \(\Delta\phi+2\pi k\) and a small neighborhood to reduce discretization error. Accuracy is sufficient due to the \(\pm0.1\) tolerance.