1. Abridged Problem Statement  
Given N distinct infinite lines on the plane (1 ≤ N ≤ 80), they partition the plane into regions—some finite, some infinite. Compute the number of finite regions and list their areas in non-decreasing order, accurate to 1e-4. Ignore any region of area ≤1e-8.

2. Detailed Editorial  

Overview  
We must find all finite faces in the planar subdivision induced by N lines and compute their areas. A robust approach uses exact arithmetic for intersections and a face-walking algorithm on the planar graph.

Steps  

1. Read Input  
   • N lines, each by two distinct integer points (x₁,y₁), (x₂,y₂).  

2. Compute All Intersections  
   • For each pair of lines i<j, check if they’re non-parallel by the determinant D = (dx₁·dy₂ – dy₁·dx₂) ≠ 0.  
   • If non-parallel, compute the intersection point P exactly as rational coordinates, store in a global set to deduplicate.  

3. Build the Planar Graph  
   • Index the unique intersection points 0…V–1.  
   • For each line, collect all intersection points that lie on it, sort these points along the line’s parameter t (or by projecting onto the line).  
   • Connect consecutive points on that line by undirected edges.  

4. Sort Neighbors by Angle  
   • For each vertex u, take its adjacency list of neighbors.  
   • Convert each neighbor v to a vector Δ = (x_v–x_u, y_v–y_u) in floating form and compute its angle θ = atan2(Δy,Δx).  
   • Sort neighbors in increasing θ order (counter-clockwise around u). This ordering lets us trace faces consistently.  

5. Face Tracing (Planar Embedding)  
   • We treat each undirected edge as two directed edges (u→v and v→u).  
   • Maintain a visited set for directed edges.  
   • For each unvisited directed edge (u→v), start a new face trace:  
     – Mark (u→v) visited; initialize cycle = [v].  
     – Repeat: at current vertex v coming from u, find the index k of u in v’s neighbor list; the previous neighbor in cyclic order (k–1 mod deg) is the next vertex w.  
     – Move to directed edge (v→w), mark visited, append w to cycle, and set (u,v) ← (v,w).  
     – Stop when we return to the starting directed edge.  
   • The collected cycle of vertices (excluding the duplicate final vertex) encloses one face.  

6. Compute Area  
   • If the cycle has <3 vertices, skip (no face).  
   • Compute the signed area via the shoelace formula. Only keep faces with positive signed area > 1e-8.  

7. Output  
   • Sort all face areas in non-decreasing order.  
   • Print K = number of faces, then each area to four decimal places.  

Complexities  
- Intersection computation: O(N²) pairs.  
- Sorting points on lines: total intersections ≤ N², sorting per line O(N log N).  
- Face tracing: each directed edge is visited once; total edges O(N²).  
- Overall O(N² log N) is acceptable for N ≤ 80.

3. C++ Solution with Detailed Comments  

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

// A rational number represented exactly as n/d in lowest terms
struct Rational {
    int64_t n, d;
    // Constructor: default 0/1 or given numerator and denominator
    Rational(int64_t nn = 0, int64_t dd = 1) : n(nn), d(dd) {
        normalize();
    }
    // Reduce to lowest terms, ensure d>0
    void normalize() {
        if (d < 0) {
            n = -n;
            d = -d;
        }
        int64_t g = gcd(abs(n), abs(d));
        n /= g; 
        d /= g;
    }
    // Comparison operators for storing in sets/maps
    bool operator==(const Rational& o) const { return n == o.n && d == o.d; }
    bool operator<(const Rational& o) const { return n * o.d < o.n * d; }
    // Arithmetic operators
    Rational operator+(const Rational& o) const {
        return Rational(n * o.d + o.n * d, d * o.d);
    }
    Rational operator-(const Rational& o) const {
        return Rational(n * o.d - o.n * d, d * o.d);
    }
    Rational operator*(const Rational& o) const {
        return Rational(n * o.n, d * o.d);
    }
    Rational operator/(const Rational& o) const {
        return Rational(n * o.d, d * o.n);
    }
};

// A 2D point with rational coordinates
struct Point {
    Rational x, y;
    Point(Rational xx = 0, Rational yy = 0) : x(xx), y(yy) {}
    bool operator==(const Point& o) const { return x == o.x && y == o.y; }
    bool operator<(const Point& o) const {
        if (!(x == o.x)) return x < o.x;
        return y < o.y;
    }
};

// A line given by two integer points (x1,y1)-(x2,y2)
struct Line {
    int64_t x1, y1, x2, y2;
};

int N;
vector<Line> lines;

// Read input: N and then each line’s two points
void read_input() {
    cin >> N;
    lines.resize(N);
    for (int i = 0; i < N; i++) {
        cin >> lines[i].x1 >> lines[i].y1
            >> lines[i].x2 >> lines[i].y2;
    }
}

void solve() {
    // 1) Compute all intersection points (rationals), collect in a set
    set<Point> all_vertices;
    for (int i = 0; i < N; i++) {
        auto &L1 = lines[i];
        int64_t dx1 = L1.x2 - L1.x1;
        int64_t dy1 = L1.y2 - L1.y1;
        for (int j = i + 1; j < N; j++) {
            auto &L2 = lines[j];
            int64_t dx2 = L2.x2 - L2.x1;
            int64_t dy2 = L2.y2 - L2.y1;
            // Determinant to check parallelism
            int64_t den = dx1 * dy2 - dy1 * dx2;
            if (den == 0) continue; // parallel or coincident → no unique intersection
            // Solve for parameter t on line1: L1.p + t*(dx1,dy1) = intersection
            int64_t num_t = (L2.x1 - L1.x1) * dy2
                          - (L2.y1 - L1.y1) * dx2;
            // Intersection coordinates = L1.p + t*(dx1,dy1)
            // We compute x_num = L1.x1*den + num_t*dx1, similarly for y
            int64_t x_num = L1.x1 * den + num_t * dx1;
            int64_t y_num = L1.y1 * den + num_t * dy1;
            // Build rational intersection point
            Point P(Rational(x_num, den), Rational(y_num, den));
            all_vertices.insert(P);
        }
    }

    // 2) Index the vertices and prepare adjacency
    int V = all_vertices.size();
    vector<Point> verts(all_vertices.begin(), all_vertices.end());
    map<Point,int> vid;
    for (int i = 0; i < V; i++) {
        vid[verts[i]] = i;
    }
    vector<vector<int>> adj(V);

    // 3) For each line, gather intersections on it, sort along the line, connect consecutive
    for (int i = 0; i < N; i++) {
        auto &L = lines[i];
        vector<pair<Rational,int>> pts; // (parameter t, vertex_id)
        int64_t dx = L.x2 - L.x1, dy = L.y2 - L.y1;
        for (int v = 0; v < V; v++) {
            // Check if vertex lies on the line via cross-product zero:
            // (P - L1) × (dx,dy) == 0
            Rational rx = verts[v].x - Rational(L.x1,1);
            Rational ry = verts[v].y - Rational(L.y1,1);
            if (rx * Rational(dy,1) == ry * Rational(dx,1)) {
                // Compute t = projection onto direction (choose dx ≠0 or dy)
                Rational t = (dx != 0)
                    ? (verts[v].x - Rational(L.x1,1)) / Rational(dx,1)
                    : (verts[v].y - Rational(L.y1,1)) / Rational(dy,1);
                pts.emplace_back(t, v);
            }
        }
        // Sort by t along the line, then connect neighbors
        sort(pts.begin(), pts.end(), [](auto &a, auto &b){
            return a.first < b.first;
        });
        for (int k = 1; k < (int)pts.size(); k++) {
            int u = pts[k-1].second;
            int w = pts[k].second;
            adj[u].push_back(w);
            adj[w].push_back(u);
        }
    }

    // 4) Sort neighbor lists by angle around each vertex
    vector<vector<int>> nbr_sorted(V);
    for (int u = 0; u < V; u++) {
        auto &nb = adj[u];
        // Remove duplicate neighbors (possible if 3+ lines meet)
        sort(nb.begin(), nb.end());
        nb.erase(unique(nb.begin(), nb.end()), nb.end());
        vector<int> order = nb;
        sort(order.begin(), order.end(), [&](int a, int b){
            // Vector from u→a
            double ax = double(verts[a].x.n)/verts[a].x.d - double(verts[u].x.n)/verts[u].x.d;
            double ay = double(verts[a].y.n)/verts[a].y.d - double(verts[u].y.n)/verts[u].y.d;
            double bx = double(verts[b].x.n)/verts[b].x.d - double(verts[u].x.n)/verts[u].x.d;
            double by = double(verts[b].y.n)/verts[b].y.d - double(verts[u].y.n)/verts[u].y.d;
            return atan2(ay,ax) < atan2(by,bx);
        });
        nbr_sorted[u] = order;
    }

    // 5) Face tracing using directed edges
    set<pair<int,int>> used_dir;
    vector<double> areas;

    for (int u = 0; u < V; u++) {
        for (int v : adj[u]) {
            if (used_dir.count({u,v})) continue;
            // Start tracing from directed edge u→v
            int su = u, sv = v;
            vector<int> cycle;
            used_dir.insert({su,sv});
            cycle.push_back(sv);
            int pu = su, pv = sv;
            while (true) {
                // In pv’s neighbor list, find index of pu
                auto &lst = nbr_sorted[pv];
                int deg = lst.size();
                int idx = find(lst.begin(), lst.end(), pu) - lst.begin();
                // Turn “right” in cyclic order: (idx - 1 + deg) % deg
                int ni = (idx - 1 + deg) % deg;
                int nv = lst[ni];
                // Move to directed edge pv→nv
                if (used_dir.count({pv,nv})) {
                    // already traced this directed edge, but we still complete the cycle
                }
                used_dir.insert({pv,nv});
                cycle.push_back(nv);
                pu = pv; pv = nv;
                if (pu == su && pv == sv) break;
            }
            // Remove the last repeated vertex
            cycle.pop_back();
            if (cycle.size() < 3) continue; // not a face
            // Compute signed area via shoelace
            double A = 0;
            int M = cycle.size();
            for (int i = 0; i < M; i++) {
                auto &P1 = verts[cycle[i]];
                auto &P2 = verts[cycle[(i+1)%M]];
                double x1 = double(P1.x.n)/P1.x.d, y1 = double(P1.y.n)/P1.y.d;
                double x2 = double(P2.x.n)/P2.x.d, y2 = double(P2.y.n)/P2.y.d;
                A += x1*y2 - x2*y1;
            }
            A *= 0.5;
            if (A > 1e-8) areas.push_back(A);
        }
    }

    // 6) Output sorted positive areas
    sort(areas.begin(), areas.end());
    cout << areas.size() << "\n" << fixed << setprecision(4);
    for (double a : areas) cout << a << "\n";
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    read_input();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  

```python
import sys
import threading
def main():
    import math
    sys.setrecursionlimit(10**7)
    data = sys.stdin.read().split()
    it = iter(data)
    N = int(next(it))
    lines = []
    for _ in range(N):
        x1,y1,x2,y2 = map(int, (next(it),next(it),next(it),next(it)))
        lines.append((x1,y1,x2,y2))

    # Compute intersections exactly as rationals (num,den) in lowest terms
    from fractions import Fraction
    pts_set = set()
    for i in range(N):
        x1,y1,x2,y2 = lines[i]
        dx1, dy1 = x2-x1, y2-y1
        for j in range(i+1, N):
            X1,Y1,X2,Y2 = lines[j]
            dx2, dy2 = X2-X1, Y2-Y1
            det = dx1*dy2 - dy1*dx2
            if det == 0:
                continue  # parallel or coincident
            # solve for parameter t on line i
            num_t = (X1-x1)*dy2 - (Y1-y1)*dx2
            # intersection = (x1 + t*dx1, y1 + t*dy1)
            # as rationals:
            t = Fraction(num_t, det)
            xi = Fraction(x1,1) + t * dx1
            yi = Fraction(y1,1) + t * dy1
            pts_set.add((xi, yi))

    # Index vertices
    verts = list(pts_set)
    V = len(verts)
    vid = {p:i for i,p in enumerate(verts)}
    adj = [[] for _ in range(V)]

    # For each line, collect intersection points on it, sort by parameter t, link consecutive
    for (x1,y1,x2,y2) in lines:
        dx, dy = x2-x1, y2-y1
        temp = []
        for idx,(xi,yi) in enumerate(verts):
            # check collinearity via cross product
            if (xi - x1)*dy == (yi - y1)*dx:
                # project to line: t = (xi-x1)/dx if dx!=0 else (yi-y1)/dy
                if dx != 0:
                    t = (xi - x1)/dx
                else:
                    t = (yi - y1)/dy
                temp.append((t, idx))
        temp.sort(key=lambda x: x[0])
        for k in range(1, len(temp)):
            u = temp[k-1][1]
            v = temp[k][1]
            adj[u].append(v)
            adj[v].append(u)

    # Sort neighbors of each vertex by angle for planar embedding
    nbr_sorted = []
    for i,(xi,yi) in enumerate(verts):
        # remove duplicates
        neigh = sorted(set(adj[i]))
        def angle(j):
            xj,yj = verts[j]
            dx, dy = float(xj-xi), float(yj-yi)
            return math.atan2(dy, dx)
        neigh.sort(key=angle)
        nbr_sorted.append(neigh)

    # Face tracing: each undirected edge → two directed edges
    used = set()
    areas = []
    for u in range(V):
        for v in adj[u]:
            if (u,v) in used:
                continue
            # start face walk from u->v
            su, sv = u, v
            used.add((su,sv))
            cycle = [sv]
            pu, pv = su, sv
            while True:
                lst = nbr_sorted[pv]
                # find index of pu in pv's neighbor list
                k = lst.index(pu)
                # next neighbor in clockwise order = (k-1) mod deg
                nv = lst[(k-1) % len(lst)]
                used.add((pv,nv))
                cycle.append(nv)
                pu, pv = pv, nv
                if pu==su and pv==sv:
                    break
            cycle.pop()  # remove repeated start
            if len(cycle) < 3:
                continue
            # compute signed area
            A = 0
            for i in range(len(cycle)):
                x1,y1 = verts[cycle[i]]
                x2,y2 = verts[cycle[(i+1)%len(cycle)]]
                A += float(x1*y2 - x2*y1)
            A *= 0.5
            if A > 1e-8:
                areas.append(A)

    areas.sort()
    # Output
    print(len(areas))
    for a in areas:
        print(f"{a:.4f}")

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

5. Compressed Editorial  

1. Compute all pairwise line intersections as exact rationals; collect unique vertices.  
2. For each line, sort intersections along it and link consecutive points to form edges.  
3. At each vertex, sort neighbors by angle to enable consistent face traversal.  
4. Trace faces by walking directed edges, always turning “right” in the sorted neighbor list.  
5. Compute shoelace area for each cycle; keep positive areas >1e-8, sort, and output.