<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

383. Caravans
Time limit per test: 4.25 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard



In this task your goal is to prey upon caravans.

There are n oases in the desert (for our purposes they are points on the plane). Sometimes caravans go from one oasis to another one. In order to prey upon them, you should predict their paths. But how to do it? The answer was given by Nomad. Caravans' velocity is constant, and they try to minimize maximal period of time outside oases. So, you can conclude, that the optimal path is a polyline. You are given several pairs of oases, and you are to output length of the maximal segment of the optimal path of a caravan which starts its way from the first oasis of the pair and ends in the second one. All oases have distinct locations and there are three oases that do not belong to one line.

Input
First line of the input contains n — amount of oases (). The following n lines describe them. Each line contains two integer numbers — xi and yi (). Next line contains one integer number q — amount of caravans (). The next q lines contain start and end oases of caravans — si and ti (1 ≤ si, ti ≤ n).

Output
Output q lengths with relative or absolute error 10-9 — one number at a line.

Example(s)
sample input
sample output
3
0 0
50 10
150 0
3
1 2
1 3
2 3
50.9901951359
100.4987562112
100.4987562112

<|response|>
## 1) Abridged problem statement

- You are given `n` distinct points in the plane (oases).
- A caravan must travel from oasis `s` to oasis `t` along a polyline that goes through some sequence of oases.
- The caravan moves at constant speed and wants to **minimize the longest straight segment** it ever travels between consecutive oases on its chosen route.
- For each of `q` queries `(s_i, t_i)`, compute this minimal possible longest segment length, with absolute or relative error ≤ `1e-9`.

Equivalently: model the oases as vertices of a complete graph with edge weights = Euclidean distances. For each query `(s, t)`, find the minimal possible value of `max edge weight on a path from s to t`.

---

## 2) Key observations

1. **Minimax path = MST path bottleneck**

   In any connected, undirected graph with positive edge weights:

   > For any two vertices `u, v`, if you take **any Minimum Spanning Tree (MST)** of the graph, then the **unique path** between `u` and `v` within that MST minimizes the maximum edge weight among all `u–v` paths in the original graph.

   So for our problem:
   - Let `G` be the complete Euclidean graph on the oases.
   - Build an MST of `G` (Euclidean MST).
   - For each query `(s, t)`, the answer is the maximum edge weight on the unique `s-t` path in the MST.

2. **We cannot build the complete graph**

   The complete graph on `n` points has `O(n^2)` edges, which is impossible to enumerate when `n` is large. We must avoid considering all pairs of points.

3. **Euclidean MST is a subgraph of the Delaunay triangulation**

   For points in 2D with Euclidean distance:
   - The **Delaunay triangulation** is a planar graph with only `O(n)` edges.
   - The Euclidean MST is a subgraph of the Delaunay triangulation.
   - So if we get the Delaunay triangulation edges, we can run Kruskal's MST algorithm on that much smaller set of edges.

4. **Voronoi diagram ↔ Delaunay triangulation**

   - The Delaunay triangulation is the planar dual of the Voronoi diagram.
   - If two sites' Voronoi cells share an edge, there is a Delaunay edge between those two points.
   - A sweep-line implementation (Fortune's algorithm) enumerates all Delaunay edges in `O(n log n)`.

5. **Parallel binary search for many minimax path queries**

   Once we have the sorted Delaunay edges, for each query `(s, t)` we need the smallest edge index `k` such that `s` and `t` become connected when we union edges `0..k` in Kruskal order. Instead of `q` independent binary searches (each rebuilding DSU in `O(|E|)`), we do parallel binary search:
   - Maintain `[low[i], high[i]]` for each query.
   - Each iteration: sweep all edges once, check connectivity at each query's current midpoint, halve intervals.
   - After `O(log |E|)` iterations, each query's answer index is `high[i] + 1`.

6. **Special case `s == t`**: answer is `0.0`.

---

## 3) Full solution approach

### Step 1: Read input
Read `n` points and `q` queries `(s_i, t_i)`, convert to 0-based indices.

### Step 2: Build Delaunay triangulation via Voronoi
Use Fortune's sweep-line Voronoi algorithm (`VoronoiDiagram` class). It:
- Rotates all points by 1 radian to avoid degenerate configurations.
- Maintains a beach line (`multiset<Arc>`) and a priority queue of site and circle events.
- Emits a Delaunay edge `(u, v)` for each pair of sites whose Voronoi cells share a border.

Complexity: `O(n log n)`.

### Step 3: Sort Delaunay edges by squared length
Sort the `m = O(n)` edges; use squared distances for comparison (actual distances only needed at output).

### Step 4: Parallel binary search over edge indices
For each query `i` (with `s_i != t_i`), find the minimal `k` such that `s_i` and `t_i` are connected using edges `0..k`.

Each iteration:
1. Group active queries by their current midpoint.
2. Sweep all edges once, unioning in DSU.
3. At each edge index `i`, answer all queries with midpoint `i`:
   - Connected → `high[qi] = i - 1`.
   - Not connected → `low[qi] = i + 1`.

After `O(log m)` iterations, `pos_i = high[i] + 1` is the answer edge index.

### Step 5: Convert index to distance
For each query `i`:
- If `s_i == t_i`, output `0.0`.
- Else output `norm(pnts[edges[pos_i].first] - pnts[edges[pos_i].second])`.

Complexity: `O((m + q) log m)` total.

---

## 4) C++ Solution

```cpp
#include <bits/stdc++.h>
// #include <coding_library/geometry/point.hpp>
// #include <coding_library/geometry/voronoi.hpp>
// #include <coding_library/graph/dsu.hpp>

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;
};

using coord_t = long double;

struct Point {
    static constexpr coord_t eps = 1e-12;

    coord_t x, y;
    Point(coord_t x = 0, coord_t y = 0) : x(x), y(y) {}

    Point operator+(const Point& p) const { return Point(x + p.x, y + p.y); }
    Point operator-(const Point& p) const { return Point(x - p.x, y - p.y); }
    Point operator*(coord_t c) const { return Point(x * c, y * c); }
    Point operator/(coord_t c) const { return Point(x / c, y / c); }

    coord_t operator*(const Point& p) const { return x * p.x + y * p.y; }
    coord_t operator^(const Point& p) const { return x * p.y - y * p.x; }

    bool operator==(const Point& p) const { return x == p.x && y == p.y; }
    bool operator!=(const Point& p) const { return x != p.x || y != p.y; }
    bool operator<(const Point& p) const {
        return x != p.x ? x < p.x : y < p.y;
    }
    bool operator>(const Point& p) const {
        return x != p.x ? x > p.x : y > p.y;
    }
    bool operator<=(const Point& p) const {
        return x != p.x ? x < p.x : y <= p.y;
    }
    bool operator>=(const Point& p) const {
        return x != p.x ? x > p.x : y >= p.y;
    }

    coord_t norm2() const { return x * x + y * y; }
    coord_t norm() const { return sqrt(norm2()); }
    coord_t angle() const { return atan2(y, x); }

    Point rotate(coord_t a) const {
        return Point(x * cos(a) - y * sin(a), x * sin(a) + y * cos(a));
    }

    Point perp() const { return Point(-y, x); }
    Point unit() const { return *this / norm(); }
    Point normal() const { return perp().unit(); }
    Point project(const Point& p) const {
        return *this * (*this * p) / norm2();
    }
    Point reflect(const Point& p) const {
        return *this * 2 * (*this * p) / norm2() - p;
    }

    friend ostream& operator<<(ostream& os, const Point& p) {
        return os << p.x << ' ' << p.y;
    }
    friend istream& operator>>(istream& is, Point& p) {
        return is >> p.x >> p.y;
    }

    friend int ccw(const Point& a, const Point& b, const Point& c) {
        coord_t v = (b - a) ^ (c - a);
        if(-eps <= v && v <= eps) {
            return 0;
        } else if(v > 0) {
            return 1;
        } else {
            return -1;
        }
    }

    friend bool point_on_segment(
        const Point& a, const Point& b, const Point& p
    ) {
        return ccw(a, b, p) == 0 && p.x >= min(a.x, b.x) - eps &&
               p.x <= max(a.x, b.x) + eps && p.y >= min(a.y, b.y) - eps &&
               p.y <= max(a.y, b.y) + eps;
    }

    friend bool point_in_triangle(
        const Point& a, const Point& b, const Point& c, const Point& p
    ) {
        int d1 = ccw(a, b, p);
        int d2 = ccw(b, c, p);
        int d3 = ccw(c, a, p);
        return (d1 >= 0 && d2 >= 0 && d3 >= 0) ||
               (d1 <= 0 && d2 <= 0 && d3 <= 0);
    }

    friend Point line_line_intersection(
        const Point& a1, const Point& b1, const Point& a2, const Point& b2
    ) {
        return a1 +
               (b1 - a1) * ((a2 - a1) ^ (b2 - a2)) / ((b1 - a1) ^ (b2 - a2));
    }

    friend bool collinear(const Point& a, const Point& b) {
        return abs(a ^ b) < eps;
    }

    friend Point circumcenter(const Point& a, const Point& b, const Point& c) {
        Point mid_ab = (a + b) / 2.0;
        Point mid_ac = (a + c) / 2.0;
        Point perp_ab = (b - a).perp();
        Point perp_ac = (c - a).perp();
        return line_line_intersection(
            mid_ab, mid_ab + perp_ab, mid_ac, mid_ac + perp_ac
        );
    }
};

class VoronoiDiagram {
  private:
    static constexpr coord_t INF = 1e100;
    static inline coord_t sweep_x;

    struct Arc {
        mutable Point p, q;
        mutable int id = 0, i;

        Arc(const Point& p, const Point& q, int i) : p(p), q(q), i(i) {}

        coord_t get_y(coord_t x) const {
            if(q.y == INF) {
                return INF;
            }
            x += Point::eps;
            Point mid = (p + q) / 2.0;
            Point dir = (p - mid).perp();
            coord_t D = (x - p.x) * (x - q.x);
            if(D < 0) {
                return -INF;
            }

            if(abs(dir.y) < Point::eps) {
                return (x < mid.x) ? -INF : INF;
            }

            return mid.y +
                   ((mid.x - x) * dir.x + sqrtl(D) * dir.norm()) / dir.y;
        }

        bool operator<(const coord_t& y) const { return get_y(sweep_x) < y; }

        bool operator<(const Arc& o) const {
            return get_y(sweep_x) < o.get_y(sweep_x);
        }
    };

    using Beach = multiset<Arc, less<>>;

    struct Event {
        coord_t x;
        int id;
        Beach::iterator it;

        Event(coord_t x, int id, Beach::iterator it) : x(x), id(id), it(it) {}

        bool operator<(const Event& e) const { return x > e.x; }
    };

    Beach beach_line;
    vector<pair<Point, int>> vertices;
    priority_queue<Event> event_queue;
    vector<pair<int, int>> edges;
    vector<bool> valid;
    int n, next_vertex_id;

    void update_vertex_event(Beach::iterator it) {
        if(it->i == -1) {
            return;
        }

        valid[-it->id] = false;
        auto prev_it = prev(it);

        if(collinear(it->q - it->p, prev_it->p - it->p)) {
            return;
        }

        it->id = --next_vertex_id;
        valid.push_back(true);

        Point center = circumcenter(it->p, it->q, prev_it->p);
        coord_t event_x = center.x + (center - it->p).norm();

        bool valid_event =
            event_x > sweep_x - Point::eps &&
            prev_it->get_y(event_x) + Point::eps > it->get_y(event_x);
        if(valid_event) {
            event_queue.push(Event(event_x, it->id, it));
        }
    }

    void add_edge(int i, int j) {
        if(i == -1 || j == -1) {
            return;
        }
        edges.push_back({vertices[i].second, vertices[j].second});
    }

    void add_point(int i) {
        Point p = vertices[i].first;

        auto split_it = beach_line.lower_bound(p.y);
        auto new_it = beach_line.insert(split_it, Arc(p, split_it->p, i));
        auto prev_it =
            beach_line.insert(new_it, Arc(split_it->p, p, split_it->i));

        add_edge(i, split_it->i);

        update_vertex_event(prev_it);
        update_vertex_event(new_it);
        update_vertex_event(split_it);
    }

    void remove_arc(Beach::iterator it) {
        auto prev_it = prev(it);
        auto next_it = next(it);

        beach_line.erase(it);
        prev_it->q = next_it->p;

        add_edge(prev_it->i, next_it->i);

        update_vertex_event(prev_it);
        update_vertex_event(next_it);
    }

  public:
    VoronoiDiagram(const vector<Point>& points, bool fix_coordinates = true) {
        n = points.size();
        vertices.resize(n);

        for(int i = 0; i < n; i++) {
            vertices[i] = {points[i], i};
        }

        if(fix_coordinates && n > 0) {
            // Rotate around center by 1.0 radians.
            for(int i = 0; i < n; i++) {
                vertices[i].first = vertices[i].first.rotate(1.0);
            }
        }

        sort(vertices.begin(), vertices.end());
    }

    vector<pair<int, int>> compute(coord_t X = 1e9) {
        edges.clear();
        beach_line.clear();
        event_queue = priority_queue<Event>();

        X *= 3;
        beach_line.insert(Arc(Point(-X, -X), Point(-X, X), -1));
        beach_line.insert(Arc(Point(-X, X), Point(INF, INF), -1));

        for(int i = 0; i < n; i++) {
            event_queue.push(Event(vertices[i].first.x, i, beach_line.end()));
        }

        next_vertex_id = 0;
        valid.assign(1, false);

        while(!event_queue.empty()) {
            Event e = event_queue.top();
            event_queue.pop();
            sweep_x = e.x;

            if(e.id >= 0) {
                add_point(e.id);
            } else if(valid[-e.id]) {
                remove_arc(e.it);
            }
        }

        return edges;
    }

    const vector<pair<int, int>>& get_edges() const { return edges; }
};

class DSU {
  public:
    int n;
    vector<int> par;
    vector<int> sz;

    DSU(int _n = 0) { init(_n); }

    void init(int _n) {
        n = _n;
        par.assign(n + 1, 0);
        sz.assign(n + 1, 0);
        for(int i = 0; i <= n; i++) {
            par[i] = i;
            sz[i] = 1;
        }
    }

    int root(int u) { return par[u] = ((u == par[u]) ? u : root(par[u])); }
    bool connected(int x, int y) { return root(x) == root(y); }

    int unite(int x, int y) {
        x = root(x), y = root(y);
        if(x == y) {
            return x;
        }
        if(sz[x] > sz[y]) {
            swap(x, y);
        }
        par[x] = y;
        sz[y] += sz[x];
        return y;
    }

    vector<vector<int>> components() {
        vector<vector<int>> comp(n + 1);
        for(int i = 0; i <= n; i++) {
            comp[root(i)].push_back(i);
        }
        return comp;
    }
};

int n, q;
vector<Point> pnts;
vector<pair<int, int>> queries;

void read() {
    cin >> n;
    pnts.resize(n);
    for(auto& p: pnts) {
        cin >> p.x >> p.y;
    }

    cin >> q;
    queries.resize(q);
    for(auto& query: queries) {
        cin >> query;
        query.first--;
        query.second--;
    }
}

void solve() {
    // The key to this problem is that we need to build the Euclidean spanning
    // tree, and then the problem ends up being computing the largest edge
    // between the two endpoints. This can be done in different ways, easiest
    // probably being parallel binary search as part of the Kruskal which we
    // anyways need to do. One corner case that isn't well described in the
    // problem statement is s[i] = t[i], where the answer should be 0.0.

    auto voronoi = VoronoiDiagram(pnts);
    auto edges = voronoi.compute();

    sort(
        edges.begin(), edges.end(),
        [&](pair<int, int> edge1, pair<int, int> edge2) {
            return (pnts[edge1.first] - pnts[edge1.second]).norm2() <
                   (pnts[edge2.first] - pnts[edge2.second]).norm2();
        }
    );

    vector<int> low(q, 0), high(q, edges.size() - 1);
    while(true) {
        bool has_queries = false;
        vector<vector<int>> queries_at_pos(edges.size());
        for(int i = 0; i < q; i++) {
            if(low[i] <= high[i]) {
                has_queries = true;

                int mid = (low[i] + high[i]) / 2;
                queries_at_pos[mid].push_back(i);
            }
        }

        if(!has_queries) {
            break;
        }

        DSU dsu(n);
        for(int i = 0; i < (int)edges.size(); i++) {
            dsu.unite(edges[i].first, edges[i].second);

            for(int query: queries_at_pos[i]) {
                if(dsu.connected(queries[query].first, queries[query].second)) {
                    high[query] = i - 1;
                } else {
                    low[query] = i + 1;
                }
            }
        }
    }

    for(int i = 0; i < q; i++) {
        if(queries[i].first == queries[i].second) {
            cout << setprecision(10) << fixed << 0.0 << endl;
        } else {
            auto edge = edges[high[i] + 1];
            cout << setprecision(10) << fixed
                 << (pnts[edge.first] - pnts[edge.second]).norm() << 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();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

---

## 5) Python implementation with detailed comments

Python does not have a Voronoi/Delaunay implementation in the standard library. The simplest practical approach is to use `scipy.spatial.Delaunay` if you're running locally. On a typical online judge, you would need to port a triangulation algorithm (complex and long), so below code assumes `scipy` is available.

If `scipy` is *not* allowed, the overall algorithm is the same, but you must replace the Delaunay part with your own implementation or switch to a C++ solution.

```python
import sys
import math
from collections import defaultdict

# Try to import Delaunay triangulation from scipy
try:
    from scipy.spatial import Delaunay
except ImportError:
    Delaunay = None  # Placeholder; see note above.


class DSU:
    """Disjoint Set Union (Union-Find) with path compression and union by size."""

    def __init__(self, n):
        self.par = list(range(n))
        self.sz = [1] * n

    def root(self, x):
        if self.par[x] != x:
            self.par[x] = self.root(self.par[x])
        return self.par[x]

    def unite(self, a, b):
        a = self.root(a)
        b = self.root(b)
        if a == b:
            return
        if self.sz[a] > self.sz[b]:
            a, b = b, a
        self.par[a] = b
        self.sz[b] += self.sz[a]

    def connected(self, a, b):
        return self.root(a) == self.root(b)


def main():
    data = sys.stdin.read().strip().split()
    it = iter(data)

    n = int(next(it))
    points = []
    for _ in range(n):
        x = float(next(it))
        y = float(next(it))
        points.append((x, y))

    q = int(next(it))
    queries = []
    for _ in range(q):
        s = int(next(it)) - 1
        t = int(next(it)) - 1
        queries.append((s, t))

    # Edge case: no or single point
    if n <= 1:
        # All answers are 0 because we can't move anywhere
        out = ["0.0000000000"] * q
        sys.stdout.write("\n".join(out))
        return

    if Delaunay is None:
        raise RuntimeError("scipy.spatial.Delaunay not available. "
                           "You need a Delaunay implementation to run this.")

    # Build Delaunay triangulation
    coords = points
    tri = Delaunay(coords)

    # Extract undirected edges from Delaunay triangles
    edge_set = set()
    for simplex in tri.simplices:
        i, j, k = simplex
        # Add the 3 undirected edges (i,j), (j,k), (k,i)
        a, b = sorted((i, j))
        c, d = sorted((j, k))
        e, f = sorted((i, k))
        edge_set.add((a, b))
        edge_set.add((c, d))
        edge_set.add((e, f))

    edges = list(edge_set)

    # Sort edges by squared Euclidean distance
    def sqdist(e):
        u, v = e
        dx = coords[u][0] - coords[v][0]
        dy = coords[u][1] - coords[v][1]
        return dx * dx + dy * dy

    edges.sort(key=sqdist)
    m = len(edges)

    # Parallel binary search over edge indices
    low = [0] * q
    high = [m - 1] * q

    # Mark queries with s == t, for which answer is 0
    same = [queries[i][0] == queries[i][1] for i in range(q)]

    while True:
        has_queries = False
        queries_at_pos = [[] for _ in range(m)]

        for i in range(q):
            if same[i]:
                continue  # skip, answer is 0
            if low[i] <= high[i]:
                has_queries = True
                mid = (low[i] + high[i]) // 2
                queries_at_pos[mid].append(i)

        if not has_queries:
            break

        dsu = DSU(n)
        # Sweep edges 0..m-1
        for ei in range(m):
            u, v = edges[ei]
            dsu.unite(u, v)

            for qi in queries_at_pos[ei]:
                s, t = queries[qi]
                if dsu.connected(s, t):
                    # They connect at or before ei → answer index <= ei
                    high[qi] = ei - 1
                else:
                    low[qi] = ei + 1

    # Construct answers
    out_lines = []
    for i in range(q):
        if same[i]:
            out_lines.append(f"{0.0:.10f}")
        else:
            idx = high[i] + 1
            u, v = edges[idx]
            dx = coords[u][0] - coords[v][0]
            dy = coords[u][1] - coords[v][1]
            dist = math.hypot(dx, dy)
            out_lines.append(f"{dist:.10f}")

    sys.stdout.write("\n".join(out_lines))


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

**Summary:**
- Model the problem as a minimax path problem on a complete Euclidean graph.
- Use the MST minimax property: the answer is the maximum edge on the MST path between the two nodes.
- Avoid `O(n^2)` edges by constructing the Euclidean MST using only Delaunay edges obtained via a Voronoi diagram (C++) or via `scipy.spatial.Delaunay` (Python).
- Answer many queries by parallel binary search combined with a Kruskal-style DSU sweep over sorted edges.
