1) Abridged problem statement
- You are given N solid circular island pieces (xi, yi, ri) in the plane and a circular ship (cx, cy, cr).
- The ship may move freely (can touch islands, but must never intersect them).
- Decide whether the ship can move arbitrarily far away from its starting position (escape to infinity). Print YES if it can, otherwise NO.

2) Detailed editorial
- Geometric reduction (Minkowski sum):
  - Expand every island radius by the ship radius cr and shift coordinates so the ship starts at the origin.
  - After this, the ship is a point at (0,0), and the obstacles are discs centered at (xi−cx, yi−cy) with radii (ri+cr). The initial position is guaranteed to be in free space.

- Key observation:
  - The origin is trapped (cannot escape to infinity) if and only if the union of the expanded discs contains a closed curve that winds around the origin. A practical sufficient/necessary certificate is a cycle of pairwise-intersecting discs that encircles the origin.

- Turning-angle graph:
  - For each disc i, compute its polar angle θi around the origin.
  - For any two discs i and j that intersect or touch (distance between centers ≤ sum of radii), consider the minimal signed angular rotation δ(i→j) needed to turn from θi to θj. Normalize δ to be in (−π, π].
  - Build a directed graph with an edge i→j of weight δ(i→j). Also add j→i with weight −δ(i→j).

- Why this works:
  - Along a chain of overlapping discs, the “view” from the origin cannot jump more than π at each step; the cumulative rotation along a closed loop around the origin must be ±2π. Thus, a loop that surrounds the origin clockwise induces a total δ-sum near −2π; counterclockwise yields +2π. Because we add both directions for each pair, any enclosing loop yields a negative cycle in at least one direction.
  - Therefore, detecting a negative cycle of magnitude at least about π is enough to conclude entrapment. The solution flags entrapment when the shortest i→i cycle sum is < −π.

- Algorithm:
  - Inflate and shift all discs.
  - For each pair (i, j), if discs overlap, compute δ(i→j) and set dist[i][j] = δ(i→j), dist[j][i] = −δ(i→j); if they don’t overlap, set dist[i][j] = +∞.
  - Run Floyd–Warshall on dist with a tiny epsilon in relaxations for numerical robustness.
  - If any dist[i][i] < −π, print NO (trapped). Otherwise, print YES (can escape).

- Correctness outline:
  - If there is a cycle of overlapping discs that winds around the origin, following the cycle in the appropriate direction yields a total δ close to −2π; Floyd–Warshall will find a negative cycle with weight < −π, thus we answer NO.
  - If there is no such enclosing cycle, there is no negative cycle with weight ≤ −π; the union of obstacles does not separate the origin from infinity, hence YES.

- Complexity:
  - Building the graph: O(N^2).
  - Floyd–Warshall: O(N^3) = 27e6 updates for N=300, which fits easily in C++ within the limits.
  - Numerical details: use EPS ~ 1e−6, normalize angles with care. The C++ below uses acos with a sign fix for negative y to recover the full [0, 2π) angle.

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 PI = acos(-1);
const double eps = 1e-6;

int n;
vector<double> x, y, rad;
double cx, cy, cr;

void read() {
    cin >> n;
    x.resize(n);
    y.resize(n);
    rad.resize(n);
    for(int i = 0; i < n; i++) {
        cin >> x[i] >> y[i] >> rad[i];
    }
    cin >> cx >> cy >> cr;
}

void solve() {
    // The idea is to check if the captain is enclosed by a cycle of overlapping
    // islands. We first transform the problem by centering at the captain's
    // position and expanding island radii by the captain's radius. This reduces
    // the captain to a point at origin.
    //
    // We then build a graph where islands are nodes. If two islands overlap (or
    // touch), we create directed edges between them with weights equal to the
    // signed angle between their centers as viewed from the origin.
    //
    // If there exists a cycle in this graph with non-zero total angle
    // (specifically, a negative cycle), it means the islands form a loop around
    // the origin, trapping the captain. We use Floyd-Warshall to find shortest
    // paths and check if any node has a negative cycle to itself.

    for(int i = 0; i < n; i++) {
        x[i] -= cx;
        y[i] -= cy;
        rad[i] += cr;
    }

    vector<vector<double>> dist(n, vector<double>(n, 0));
    for(int i = 0; i < n; i++) {
        for(int j = i + 1; j < n; j++) {
            double center_dist_sq =
                (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
            double sum_rad_sq = (rad[i] + rad[j]) * (rad[i] + rad[j]);

            if(center_dist_sq + eps > sum_rad_sq) {
                dist[i][j] = 1e18;
                dist[j][i] = 1e18;
            } else {
                double alpha = acos(x[i] / sqrt(x[i] * x[i] + y[i] * y[i]));
                double beta = acos(x[j] / sqrt(x[j] * x[j] + y[j] * y[j]));
                if(y[i] < -eps) {
                    alpha = 2 * PI - alpha;
                }
                if(y[j] < -eps) {
                    beta = 2 * PI - beta;
                }

                double angle_diff = beta - alpha;
                if(angle_diff < 0) {
                    angle_diff += 2 * PI;
                }

                if(angle_diff > PI) {
                    angle_diff -= 2 * PI;
                }
                dist[i][j] = angle_diff;
                dist[j][i] = -angle_diff;
            }
        }
    }

    for(int k = 0; k < n; k++) {
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j] + eps);
            }
        }
    }

    for(int i = 0; i < n; i++) {
        if(dist[i][i] < -PI) {
            cout << "NO" << endl;
            return;
        }
    }

    cout << "YES" << 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;
}
```

4) Python solution (same idea) with detailed comments
```python
import sys
import math

def can_escape(circles, ship):
    # circles: list of (x, y, r)
    # ship: (cx, cy, cr)
    cx, cy, cr = ship
    n = len(circles)

    # Shift centers so the ship's center is at the origin and
    # inflate radii by the ship radius (Minkowski sum).
    x = [0.0] * n
    y = [0.0] * n
    R = [0.0] * n
    for i, (xi, yi, ri) in enumerate(circles):
        x[i] = xi - cx
        y[i] = yi - cy
        R[i] = ri + cr

    # Prepare angle for each center using atan2 (robust for all quadrants).
    # If a center is (0, 0) and R[i] > 0, that would imply the origin is inside a disc,
    # which contradicts "ship starts in free area". We keep the code robust anyway.
    theta = [math.atan2(y[i], x[i]) % (2.0 * math.pi) for i in range(n)]

    # Build directed graph weights:
    # w[i][j] = minimal signed angle to rotate from theta[i] to theta[j] if discs i and j intersect, else +inf.
    INF = 1e100
    w = [[INF] * n for _ in range(n)]
    for i in range(n):
        w[i][i] = 0.0

    # Epsilon used both in intersection tests and relaxations
    EPS = 1e-6
    for i in range(n):
        xi, yi, ri = x[i], y[i], R[i]
        for j in range(i + 1, n):
            xj, yj, rj = x[j], y[j], R[j]
            # Squared distance between centers
            dx = xi - xj
            dy = yi - yj
            center_dist_sq = dx * dx + dy * dy
            sum_rad = ri + rj
            sum_rad_sq = sum_rad * sum_rad

            # If they intersect/touch: center_dist <= sum_rad (+tiny tolerance).
            if center_dist_sq + EPS <= sum_rad_sq:
                # Compute minimal signed angle difference in (−π, π]
                diff = theta[j] - theta[i]
                # Normalize into (−π, π]
                diff = (diff + math.pi) % (2.0 * math.pi) - math.pi
                w[i][j] = diff
                w[j][i] = -diff
            else:
                # Not intersecting: leave as INF
                pass

    # Floyd–Warshall to find all-pairs shortest paths in the angle graph.
    # If a cycle wraps around the origin clockwise, there exists k with w[k][k] < −π after closure.
    for k in range(n):
        wk = w[k]
        for i in range(n):
            wik = w[i][k]
            if wik >= INF:
                continue
            wi = w[i]
            # Relax i->j via k
            t = wik  # w[i][k]
            for j in range(n):
                val = t + wk[j] + EPS
                if val < wi[j]:
                    wi[j] = val

    # Detect a sufficiently negative cycle (enclosing the origin)
    for i in range(n):
        if w[i][i] < -math.pi:
            return False  # cannot escape

    return True  # can escape


def main():
    data = sys.stdin.read().strip().split()
    it = iter(data)
    try:
        n = int(next(it))
    except StopIteration:
        return
    circles = []
    for _ in range(n):
        # Read xi, yi, ri as floats
        xi = float(next(it)); yi = float(next(it)); ri = float(next(it))
        circles.append((xi, yi, ri))
    cx = float(next(it)); cy = float(next(it)); cr = float(next(it))
    print("YES" if can_escape(circles, (cx, cy, cr)) else "NO")

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

5) Compressed editorial
- Inflate each island radius by the ship radius and shift coordinates so the ship becomes a point at the origin.
- Build a graph on discs: connect i and j if their inflated discs intersect. For each such pair, compute the minimal signed angular difference δ ∈ (−π, π] from the angle of ci to cj (as seen from the origin). Add directed edges i→j with weight δ and j→i with weight −δ.
- Run Floyd–Warshall on this graph. If any dist[i][i] < −π after closure, there is a cycle whose total rotation is about −2π, i.e., an enclosing loop around the origin; answer NO. Otherwise answer YES.
- Complexity: O(N^3) with N ≤ 300. Use small eps for numeric stability.
