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

```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;
};

struct Rational {
    int64_t n, d;
    Rational(int64_t _n = 0, int64_t _d = 1) : n(_n), d(_d) { normalize(); }

    void normalize() {
        if(d < 0) {
            n = -n;
            d = -d;
        }

        int64_t g = gcd(abs(n), abs(d));
        n /= g;
        d /= g;
    }

    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; }

    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);
    }
};

struct Point {
    Rational x, y;
    Point(Rational _x = 0, Rational _y = 0) : x(_x), y(_y) {}

    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;
    }
};

struct Line {
    int64_t x1, y1, x2, y2;
};

int N;
vector<Line> lines;

void read() {
    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() {
    // Build the planar arrangement of the N lines and report the areas of all
    // bounded faces, using exact rational coordinates for the vertices.
    //
    // Vertices are all pairwise intersection points (computed exactly as
    // Rational x, y). For each line we collect the vertices lying on it,
    // sort them along the line, and add graph edges between consecutive ones,
    // giving the arrangement graph. To enumerate faces we do a planar face
    // traversal: at every vertex the incident edges are sorted by angle, and
    // walking a directed edge (u, v) we always turn to the previous neighbour
    // (clockwise) around v. Each directed edge belongs to exactly one face, so
    // marking used directed edges enumerates every face once. A face is bounded
    // iff its signed area (shoelace) is positive; we collect those areas,
    // discard ones below 1e-8, and print them sorted.

    set<Point> all_vertices;
    for(int i = 0; i < N; i++) {
        for(int j = i + 1; j < N; j++) {
            int64_t dx1 = lines[i].x2 - lines[i].x1;
            int64_t dy1 = lines[i].y2 - lines[i].y1;
            int64_t dx2 = lines[j].x2 - lines[j].x1;
            int64_t dy2 = lines[j].y2 - lines[j].y1;
            int64_t den = dx1 * dy2 - dy1 * dx2;
            if(den == 0) {
                continue;
            }

            int64_t num_t = (lines[j].x1 - lines[i].x1) * dy2 -
                            (lines[j].y1 - lines[i].y1) * dx2;
            int64_t x_num = lines[i].x1 * den + num_t * dx1;
            int64_t y_num = lines[i].y1 * den + num_t * dy1;
            Point p(Rational(x_num, den), Rational(y_num, den));
            all_vertices.insert(p);
        }
    }

    vector<Point> verts(all_vertices.begin(), all_vertices.end());
    int n = verts.size();
    vector<vector<int>> adj(n);
    map<Point, int> point_to_id;
    for(int i = 0; i < n; i++) {
        point_to_id[verts[i]] = i;
    }

    for(int i = 0; i < N; i++) {
        set<Point> on_line;
        for(int j = 0; j < N; j++) {
            if(i == j) {
                continue;
            }

            int64_t dx1 = lines[i].x2 - lines[i].x1;
            int64_t dy1 = lines[i].y2 - lines[i].y1;
            int64_t dx2 = lines[j].x2 - lines[j].x1;
            int64_t dy2 = lines[j].y2 - lines[j].y1;
            int64_t den = dx1 * dy2 - dy1 * dx2;
            if(den == 0) {
                continue;
            }

            int64_t num_t = (lines[j].x1 - lines[i].x1) * dy2 -
                            (lines[j].y1 - lines[i].y1) * dx2;
            int64_t x_num = lines[i].x1 * den + num_t * dx1;
            int64_t y_num = lines[i].y1 * den + num_t * dy1;
            Point p(Rational(x_num, den), Rational(y_num, den));
            on_line.insert(p);
        }

        vector<Point> pts(on_line.begin(), on_line.end());
        for(size_t k = 0; k + 1 < pts.size(); k++) {
            int a = point_to_id[pts[k]];
            int b = point_to_id[pts[k + 1]];
            adj[a].push_back(b);
            adj[b].push_back(a);
        }
    }

    vector<vector<int>> sorted_neighbors(n);
    for(int i = 0; i < n; i++) {
        vector<int> nb = adj[i];
        sort(nb.begin(), nb.end(), [&](int a, int b) {
            Rational dx1 = verts[a].x - verts[i].x;
            Rational dy1 = verts[a].y - verts[i].y;
            double xx1 = (double)dx1.n / dx1.d;
            double yy1 = (double)dy1.n / dy1.d;
            double ang1 = atan2(yy1, xx1);
            Rational dx2 = verts[b].x - verts[i].x;
            Rational dy2 = verts[b].y - verts[i].y;
            double xx2 = (double)dx2.n / dx2.d;
            double yy2 = (double)dy2.n / dy2.d;
            double ang2 = atan2(yy2, xx2);
            return ang1 < ang2;
        });

        sorted_neighbors[i] = nb;
    }

    set<pair<int, int>> used_directed;
    vector<double> areas;
    for(int i = 0; i < n; i++) {
        for(int j: adj[i]) {
            if(used_directed.count({i, j})) {
                continue;
            }

            vector<int> cycle;
            int curr_u = i;
            int curr_v = j;
            used_directed.insert({curr_u, curr_v});
            cycle.push_back(curr_v);
            while(true) {
                int deg = sorted_neighbors[curr_v].size();
                int k = -1;
                for(int s = 0; s < deg; s++) {
                    if(sorted_neighbors[curr_v][s] == curr_u) {
                        k = s;
                        break;
                    }
                }

                int next_k = (k - 1 + deg) % deg;
                int next_v = sorted_neighbors[curr_v][next_k];
                used_directed.insert({curr_v, next_v});
                curr_u = curr_v;
                curr_v = next_v;
                cycle.push_back(curr_v);
                if(curr_v == j) {
                    break;
                }
            }

            cycle.pop_back();
            int m = cycle.size();
            if(m < 3) {
                continue;
            }

            double signed_area = 0.0;
            for(int kk = 0; kk < m; kk++) {
                int aa = cycle[kk];
                int bb = cycle[(kk + 1) % m];
                double x1 = (double)verts[aa].x.n / verts[aa].x.d;
                double y1 = (double)verts[aa].y.n / verts[aa].y.d;
                double x2 = (double)verts[bb].x.n / verts[bb].x.d;
                double y2 = (double)verts[bb].y.n / verts[bb].y.d;
                signed_area += x1 * y2 - x2 * y1;
            }

            if(signed_area > 1e-8) {
                areas.push_back(signed_area / 2.0);
            }
        }
    }

    sort(areas.begin(), areas.end());
    cout << areas.size() << '\n';
    cout << fixed << setprecision(4);
    for(double ar: areas) {
        cout << ar << '\n';
    }
}

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 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.