<|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.

373. Carlsson vs. Winnie-the-Pooh
Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



Carlsson and Winnie the Pooh are eating one pizza, which is a circle of radius R. They divided the pizza into some pieces by N straight lines. They eat pizza in a following way. Any person takes a piece and eats it, then takes the next piece, eats it and so on. They eat pieces with the same speed. The time of eating a piece is proportional to its area. If at any moment of time they are both trying to take a piece simultaneously, Carlsson does his selection first. Your goal is to write the program, which determines the amounts of pizza eaten by each side, if it is known that both sides try to maximize their total amount of eaten pizza.

Input
The first line of the input contains N (1 ≤ N≤ 4) and R (1 ≤ R≤ 100). The center of the pizza is located at (0, 0). Each of the next N lines describes one line with three integers a, b and c (-1000 ≤ a, b, c≤ 1000). These numbers correspond to equation ax + by + c = 0. It is guaranteed that each line intersects the edge of pizza in exactly two points.

Output
Output the area of pizza eaten by Carlsson and by Winnie the Pooh. The testing program checks whether the answer is within 10-4 of the right answer.

Example(s)
sample input
sample output
1 100
1 0 0
15707.9632679490 15707.9632679490

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

A pizza is a disk of radius **R** centered at (0,0). It is cut by **N** straight lines (1 ≤ N ≤ 4), each intersects the circle in exactly two points, producing several pieces.

Carlsson and Winnie-the-Pooh repeatedly:
1) take one remaining piece,
2) eat it; eating time is proportional to the piece area (same speed for both),
3) when someone finishes, they immediately take another remaining piece.

If both become free at the same time, **Carlsson chooses first**.

Both play optimally to maximize their own total eaten area. Output final areas eaten by Carlsson and Pooh (error ≤ 1e-4).

---

## 2) Key observations

1. **Eating time equals total area already taken.**  
   If Carlsson has eaten area `tc` so far and Pooh has eaten `tp`, then at any moment their “busy time” is exactly these totals.

2. **Who chooses next is determined by `tc` vs `tp`, not alternating turns.**  
   - If `tc <= tp` ⇒ Carlsson is free earlier (or tie) ⇒ Carlsson picks next.  
   - Else Pooh picks next.

3. **Game is not greedy.**  
   Taking a huge piece can delay your next pick and allow the opponent to take multiple smaller pieces.

4. **N ≤ 4 ⇒ number of pieces is very small.**  
   A set of N lines creates at most `1 + N(N+1)/2` regions in the plane (≤ 11 for N=4). Intersecting with the disk still yields O(11) pieces.  
   Therefore we can compute all piece areas and do a full minimax over subsets.

5. **We must compute exact areas of all faces formed by chords + circle boundary.**  
   Use a planar graph / half-edge (DCEL-like) traversal:
   - vertices: all line–circle intersections and line–line intersections inside the disk
   - edges: chord segments between consecutive vertices on a line + circle arcs between consecutive boundary vertices
   - traverse faces via half-edge `next` pointers
   - compute area via Green’s theorem (segments + arc contributions)

---

## 3) Full solution approach

### Part A — Build the arrangement inside the circle and extract piece areas

**A1) Collect vertices**
- For each line `ax + by + c = 0`, compute its two intersection points with the circle `x^2 + y^2 = R^2`.
- For each pair of lines, compute their intersection point; keep it only if it lies inside/on the disk.
- Deduplicate close points (floating precision).

**A2) Create edges**
- For each line:
  - collect all vertices that lie on it (its two circle endpoints + interior line-line intersections),
  - sort them along the line,
  - connect consecutive points with straight segment edges (chord pieces).
- For the circle boundary:
  - collect all circle intersection vertices,
  - sort them by polar angle,
  - connect consecutive vertices with **circular arc** edges in CCW order.

Represent each undirected edge as two directed **half-edges** with `twin` pointers.

**A3) Link half-edges into faces (DCEL rule)**
At each vertex, sort outgoing half-edges by direction angle (CCW).  
For a half-edge `h`, define:
> `next(h)` = the outgoing half-edge at `to(h)` that is immediately **clockwise** from `twin(h)`.

This ensures face traversal keeps the face on the left side.

Walking `h -> next(h) -> next(next(h)) ...` gives one face boundary.

**A4) Compute face areas**
Use Green’s theorem by summing contributions of each boundary element:

- For a **segment** from `P` to `Q`:  
  `2A += cross(P, Q) = Px*Qy - Py*Qx`
- For a **circle arc** of radius `R` centered at origin from angle `θ1` to `θ2` along the arc direction:  
  `2A += R^2 * Δθ` (signed), where `Δθ` is adjusted into `(0, 2π]` for CCW and `[-2π, 0)` for CW.

Faces inside the disk come out with **positive** signed area; the exterior face is negative.  
Collect all `A > 0` as pizza piece areas.

---

### Part B — Solve the optimal game by minimax over subsets

Let `areas[]` be all piece areas, `p = len(areas)`.

State during play:
- `mask`: bitmask of already taken pieces
- `tc`: Carlsson total eaten area so far (also time spent)
- `tp`: Pooh total eaten area so far
- objective: maximize Carlsson’s final total area

Turn rule:
- if `tc <= tp` (tie to Carlsson): Carlsson chooses next
- else Pooh chooses next

Minimax recursion:
- terminal: all taken ⇒ return Carlsson total so far
- Carlsson node: choose remaining piece i maximizing result
- Pooh node: choose remaining piece i minimizing Carlsson’s result

Because `p ≤ ~11`, brute force (with alpha-beta pruning) is fast.

Finally:
- `carl = minimax(...)`
- `pooh = total_area - carl`

---

## 4) C++ implementation (detailed comments)

```cpp
#include <bits/stdc++.h>
using namespace std;

/*
  Solution outline:
  1) Build arrangement of N<=4 lines clipped to a circle of radius R.
     Use half-edge structure to enumerate all faces (pizza pieces).
  2) Compute each face area (segments + circular arcs).
  3) Run minimax with alpha-beta: next player is the one with smaller
     accumulated eaten area (time), tie -> Carlsson.

  Complexity: geometry is tiny; number of pieces <= ~11; minimax over subsets is fine.
*/

struct Point {
    static constexpr double EPS = 1e-9;
    static inline const double PI = acos(-1.0);

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

    Point operator+(const Point& o) const { return {x+o.x, y+o.y}; }
    Point operator-(const Point& o) const { return {x-o.x, y-o.y}; }
    Point operator*(double k) const { return {x*k, y*k}; }
    Point operator/(double k) const { return {x/k, y/k}; }

    double norm() const { return hypot(x,y); }
};

static double cross(const Point& a, const Point& b) {
    return a.x*b.y - a.y*b.x;
}

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

    int N;
    double R;
    cin >> N >> R;

    vector<array<double,3>> L(N);
    for (int i=0;i<N;i++) cin >> L[i][0] >> L[i][1] >> L[i][2];

    // ---- geometry helpers ----

    // For line ax+by+c=0: return its two intersection points with circle x^2+y^2=R^2.
    auto line_circle_pts = [&](int i) -> array<Point,2> {
        double a=L[i][0], b=L[i][1], c=L[i][2];
        double n2 = a*a + b*b;

        // Foot of perpendicular from origin to the line: (-a*c/n2, -b*c/n2)
        Point foot(-a*c/n2, -b*c/n2);

        // Distance^2 from foot to intersection along the line direction
        double h2 = R*R - (c*c)/n2;
        h2 = max(0.0, h2);                 // numeric guard
        double h = sqrt(h2);

        // Unit direction along the line: perpendicular to normal (a,b) is (-b,a)
        Point dir(-b, a);
        dir = dir / sqrt(n2);

        return { foot + dir*h, foot - dir*h };
    };

    // Intersection of two infinite lines; keep only if inside/on the disk.
    auto line_line_pt = [&](int i, int j) -> optional<Point> {
        double a1=L[i][0], b1=L[i][1], c1=L[i][2];
        double a2=L[j][0], b2=L[j][1], c2=L[j][2];
        double d = a1*b2 - a2*b1;
        if (fabs(d) < Point::EPS) return nullopt;  // parallel or nearly

        // Solve:
        // a1 x + b1 y = -c1
        // a2 x + b2 y = -c2
        double x = (-c1*b2 + b1*c2)/d;
        double y = (-a1*c2 + a2*c1)/d;
        Point p(x,y);

        if (p.norm() > R + 1e-9) return nullopt;
        return p;
    };

    // ---- build vertices with dedup ----
    vector<Point> V;

    auto add_vertex = [&](const Point& p) -> int {
        for (int i=0;i<(int)V.size();i++) {
            if (hypot(V[i].x - p.x, V[i].y - p.y) < 1e-7) return i;
        }
        V.push_back(p);
        return (int)V.size()-1;
    };

    // ---- half-edge structure ----
    struct HalfEdge {
        int from, to;
        bool is_arc;   // false: segment, true: circle arc
        bool ccw;      // for arcs: direction along circle
        int twin;
        int next;
    };
    vector<HalfEdge> HE;

    auto add_edge = [&](int u, int v, bool is_arc, bool ccw_uv) {
        int h = (int)HE.size();
        HE.push_back({u, v, is_arc, ccw_uv, h+1, -1});
        HE.push_back({v, u, is_arc, !ccw_uv, h, -1});
    };

    // ---- chord segments on each line ----
    for (int i=0;i<N;i++) {
        vector<int> on_line;

        // endpoints with circle
        auto cp = line_circle_pts(i);
        on_line.push_back(add_vertex(cp[0]));
        on_line.push_back(add_vertex(cp[1]));

        // intersections with other lines
        for (int j=0;j<N;j++) if (j!=i) {
            auto p = line_line_pt(i,j);
            if (p) on_line.push_back(add_vertex(*p));
        }

        // sort along line direction using dot with (-b,a)
        double a=L[i][0], b=L[i][1];
        sort(on_line.begin(), on_line.end(), [&](int u, int v){
            double du = (-b)*V[u].x + a*V[u].y;
            double dv = (-b)*V[v].x + a*V[v].y;
            return du < dv;
        });

        on_line.erase(unique(on_line.begin(), on_line.end()), on_line.end());

        // connect consecutive points with straight edges
        for (int k=0;k+1<(int)on_line.size();k++) {
            add_edge(on_line[k], on_line[k+1], false, false);
        }
    }

    // ---- circle boundary arcs ----
    vector<int> circle_vs;
    for (int i=0;i<N;i++) {
        auto cp = line_circle_pts(i);
        circle_vs.push_back(add_vertex(cp[0]));
        circle_vs.push_back(add_vertex(cp[1]));
    }

    auto angle0_2pi = [&](int vid) {
        double a = atan2(V[vid].y, V[vid].x);
        if (a < 0) a += 2*Point::PI;
        return a;
    };

    sort(circle_vs.begin(), circle_vs.end(), [&](int u, int v){
        return angle0_2pi(u) < angle0_2pi(v);
    });
    circle_vs.erase(unique(circle_vs.begin(), circle_vs.end()), circle_vs.end());

    for (int k=0;k<(int)circle_vs.size();k++) {
        int u = circle_vs[k];
        int v = circle_vs[(k+1) % circle_vs.size()];
        // arc along boundary is CCW from u to v
        add_edge(u, v, true, true);
    }

    // direction vector for sorting outgoing half-edges at a vertex
    auto outgoing_dir = [&](int h) -> Point {
        Point p = V[HE[h].from];
        if (!HE[h].is_arc) {
            return V[HE[h].to] - p; // segment direction
        } else {
            // tangent direction to circle at p: CCW tangent is (-y,x)
            if (HE[h].ccw) return Point(-p.y, p.x);
            else          return Point(p.y, -p.x);
        }
    };

    // gather outgoing half-edges for each vertex
    int VN = (int)V.size();
    vector<vector<int>> out(VN);
    for (int h=0;h<(int)HE.size();h++) out[HE[h].from].push_back(h);

    // sort outgoing by direction angle CCW
    for (auto &lst : out) {
        sort(lst.begin(), lst.end(), [&](int ha, int hb){
            Point da = outgoing_dir(ha);
            Point db = outgoing_dir(hb);
            return atan2(da.y, da.x) < atan2(db.y, db.x);
        });
    }

    // position of each half-edge in out[from]
    vector<int> pos(HE.size());
    for (int v=0;v<VN;v++) {
        for (int i=0;i<(int)out[v].size();i++) pos[out[v][i]] = i;
    }

    // set next pointers:
    // next(h) is the outgoing edge at vertex to(h) that is immediately clockwise from twin(h)
    for (int h=0;h<(int)HE.size();h++) {
        int t = HE[h].twin;
        int v = HE[t].from; // equals to(h)
        int deg = (int)out[v].size();
        HE[h].next = out[v][(pos[t] - 1 + deg) % deg];
    }

    // ---- traverse faces, compute signed area, keep positive ones ----
    vector<char> used(HE.size(), 0);
    vector<double> areas;

    for (int h0=0;h0<(int)HE.size();h0++) if (!used[h0]) {
        double area2 = 0.0;
        int cur = h0;

        while (!used[cur]) {
            used[cur] = 1;
            Point A = V[HE[cur].from];
            Point B = V[HE[cur].to];

            if (!HE[cur].is_arc) {
                // segment contribution to 2*area
                area2 += cross(A, B);
            } else {
                // arc contribution: R^2 * dtheta (signed by direction)
                double th1 = atan2(A.y, A.x);
                double th2 = atan2(B.y, B.x);
                double dth = th2 - th1;

                if (HE[cur].ccw) {
                    // want dth in (0, 2pi]
                    if (dth <= 1e-12) dth += 2*Point::PI;
                } else {
                    // want dth in [-2pi, 0)
                    if (dth >= -1e-12) dth -= 2*Point::PI;
                }
                area2 += R*R * dth;
            }

            cur = HE[cur].next;
        }

        double A = area2 / 2.0;
        if (A > 1e-9) areas.push_back(A); // pizza pieces
    }

    // ---- minimax game on piece areas ----
    int P = (int)areas.size();
    double total = 0;
    for (double a : areas) total += a;

    function<double(int,double,double,double,double,double)> dfs =
        [&](int mask, double tc, double tp, double csf, double alpha, double beta) -> double {
            if (mask == (1<<P) - 1) return csf;

            bool carl_turn = (tc <= tp + 1e-9); // tie -> Carlsson

            if (carl_turn) {
                double best = -1e100;
                for (int i=0;i<P;i++) if (!(mask&(1<<i))) {
                    double v = dfs(mask|(1<<i), tc+areas[i], tp, csf+areas[i], alpha, beta);
                    best = max(best, v);
                    alpha = max(alpha, best);
                    if (alpha >= beta) break; // prune
                }
                return best;
            } else {
                double best = 1e100;
                for (int i=0;i<P;i++) if (!(mask&(1<<i))) {
                    double v = dfs(mask|(1<<i), tc, tp+areas[i], csf, alpha, beta);
                    best = min(best, v);
                    beta = min(beta, best);
                    if (alpha >= beta) break; // prune
                }
                return best;
            }
        };

    double carl = dfs(0, 0, 0, 0, -1e100, 1e100);
    double pooh = total - carl;

    cout << fixed << setprecision(10) << carl << " " << pooh << "\n";
    return 0;
}
```

---

## 5) Python implementation (detailed comments)

> Note: This mirrors the C++ logic, but the original time limit (0.25s) is typically C++-oriented. Python may be borderline on some judges; still, with N ≤ 4 it’s often fine.

```python
import sys, math
from typing import List, Tuple, Optional

EPS = 1e-9
PI = math.pi

class Point:
    __slots__ = ("x", "y")
    def __init__(self, x: float = 0.0, y: float = 0.0):
        self.x = x
        self.y = y
    def __add__(self, o): return Point(self.x + o.x, self.y + o.y)
    def __sub__(self, o): return Point(self.x - o.x, self.y - o.y)
    def __mul__(self, k: float): return Point(self.x * k, self.y * k)
    def __truediv__(self, k: float): return Point(self.x / k, self.y / k)
    def norm(self) -> float: return math.hypot(self.x, self.y)

def cross(a: Point, b: Point) -> float:
    return a.x * b.y - a.y * b.x

class HalfEdge:
    __slots__ = ("frm", "to", "is_arc", "ccw", "twin", "nxt")
    def __init__(self, frm: int, to: int, is_arc: bool, ccw: bool, twin: int):
        self.frm = frm
        self.to = to
        self.is_arc = is_arc
        self.ccw = ccw
        self.twin = twin
        self.nxt = -1

def solve() -> None:
    data = sys.stdin.read().strip().split()
    if not data:
        return
    it = iter(data)
    N = int(next(it))
    R = float(next(it))

    lines: List[Tuple[float, float, float]] = []
    for _ in range(N):
        a = float(next(it)); b = float(next(it)); c = float(next(it))
        lines.append((a, b, c))

    # line-circle intersections
    def line_circle_pts(i: int) -> Tuple[Point, Point]:
        a, b, c = lines[i]
        n2 = a*a + b*b
        foot = Point(-a*c/n2, -b*c/n2)
        h2 = R*R - (c*c)/n2
        if h2 < 0: h2 = 0.0
        h = math.sqrt(h2)
        dirv = Point(-b, a) / math.sqrt(n2)
        return foot + dirv*h, foot - dirv*h

    # line-line intersection if inside disk
    def line_line_pt(i: int, j: int) -> Optional[Point]:
        a1, b1, c1 = lines[i]
        a2, b2, c2 = lines[j]
        d = a1*b2 - a2*b1
        if abs(d) < EPS:
            return None
        x = (-c1*b2 + b1*c2) / d
        y = (-a1*c2 + a2*c1) / d
        p = Point(x, y)
        if p.norm() > R + 1e-9:
            return None
        return p

    # dedup vertices
    verts: List[Point] = []
    def add_vertex(p: Point) -> int:
        for idx, q in enumerate(verts):
            if math.hypot(q.x - p.x, q.y - p.y) < 1e-7:
                return idx
        verts.append(p)
        return len(verts) - 1

    # half-edges
    hes: List[HalfEdge] = []
    def add_edge(u: int, v: int, is_arc: bool, ccw_uv: bool) -> None:
        h = len(hes)
        hes.append(HalfEdge(u, v, is_arc, ccw_uv, h+1))
        hes.append(HalfEdge(v, u, is_arc, not ccw_uv, h))

    # chord segments
    for i in range(N):
        on_line: List[int] = []
        p1, p2 = line_circle_pts(i)
        on_line.append(add_vertex(p1))
        on_line.append(add_vertex(p2))

        for j in range(N):
            if j == i: continue
            p = line_line_pt(i, j)
            if p is not None:
                on_line.append(add_vertex(p))

        a, b, _ = lines[i]
        on_line.sort(key=lambda vid: (-b)*verts[vid].x + a*verts[vid].y)

        # unique vertex indices
        uniq = []
        last = None
        for vid in on_line:
            if last is None or vid != last:
                uniq.append(vid)
            last = vid
        on_line = uniq

        for k in range(len(on_line)-1):
            add_edge(on_line[k], on_line[k+1], is_arc=False, ccw_uv=False)

    # boundary arcs
    circle_vs: List[int] = []
    for i in range(N):
        p1, p2 = line_circle_pts(i)
        circle_vs.append(add_vertex(p1))
        circle_vs.append(add_vertex(p2))

    def angle_0_2pi(vid: int) -> float:
        a = math.atan2(verts[vid].y, verts[vid].x)
        if a < 0: a += 2*PI
        return a

    circle_vs.sort(key=angle_0_2pi)
    uniq = []
    last = None
    for vid in circle_vs:
        if last is None or vid != last:
            uniq.append(vid)
        last = vid
    circle_vs = uniq

    m = len(circle_vs)
    for k in range(m):
        add_edge(circle_vs[k], circle_vs[(k+1) % m], is_arc=True, ccw_uv=True)

    # direction for outgoing half-edge sorting
    def outgoing_dir(h: int) -> Point:
        p = verts[hes[h].frm]
        if not hes[h].is_arc:
            return verts[hes[h].to] - p
        # tangent direction
        if hes[h].ccw:
            return Point(-p.y, p.x)
        else:
            return Point(p.y, -p.x)

    out: List[List[int]] = [[] for _ in range(len(verts))]
    for h in range(len(hes)):
        out[hes[h].frm].append(h)

    for v in range(len(verts)):
        out[v].sort(key=lambda h: math.atan2(outgoing_dir(h).y, outgoing_dir(h).x))

    pos = [0]*len(hes)
    for v in range(len(verts)):
        for i, h in enumerate(out[v]):
            pos[h] = i

    # next pointers
    for h in range(len(hes)):
        t = hes[h].twin
        v = hes[t].frm
        deg = len(out[v])
        hes[h].nxt = out[v][(pos[t] - 1) % deg]  # clockwise from twin

    # face traversal + area
    used = [False]*len(hes)
    areas: List[float] = []
    for h0 in range(len(hes)):
        if used[h0]:
            continue
        area2 = 0.0
        cur = h0
        while not used[cur]:
            used[cur] = True
            A = verts[hes[cur].frm]
            B = verts[hes[cur].to]

            if not hes[cur].is_arc:
                area2 += cross(A, B)
            else:
                th1 = math.atan2(A.y, A.x)
                th2 = math.atan2(B.y, B.x)
                dth = th2 - th1
                if hes[cur].ccw:
                    if dth <= 1e-12: dth += 2*PI
                else:
                    if dth >= -1e-12: dth -= 2*PI
                area2 += R*R * dth

            cur = hes[cur].nxt

        A = area2 / 2.0
        if A > 1e-9:
            areas.append(A)

    total = sum(areas)
    P = len(areas)

    # minimax with alpha-beta
    sys.setrecursionlimit(10000)

    def dfs(mask: int, tc: float, tp: float, csf: float, alpha: float, beta: float) -> float:
        if mask == (1 << P) - 1:
            return csf

        carl_turn = (tc <= tp + 1e-9)  # tie to Carlsson

        if carl_turn:
            best = -1e100
            for i in range(P):
                if mask & (1 << i): 
                    continue
                v = dfs(mask | (1 << i), tc + areas[i], tp, csf + areas[i], alpha, beta)
                if v > best: best = v
                if best > alpha: alpha = best
                if alpha >= beta: break
            return best
        else:
            best = 1e100
            for i in range(P):
                if mask & (1 << i):
                    continue
                v = dfs(mask | (1 << i), tc, tp + areas[i], csf, alpha, beta)
                if v < best: best = v
                if best < beta: beta = best
                if alpha >= beta: break
            return best

    carl = dfs(0, 0.0, 0.0, 0.0, -1e100, 1e100)
    pooh = total - carl
    print(f"{carl:.10f} {pooh:.10f}")

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

---

If you want, I can also provide a lighter geometry alternative (explicitly enumerating regions via angular sorting and polygon clipping) but the half-edge face-walk above is the most systematic and reliable for mixed line segments + circular arcs.