1. Abridged Problem Statement  
Given an integer n and two n×n matrices top and left, where  
- left[i][j] = number of entries in row i to the left of (i,j) that are larger than A[i][j],  
- top[i][j]  = number of entries in column j above (i,j) that are larger than A[i][j],  
reconstruct any n×n permutation matrix A of 1…n² matching these counts, or print 0 if impossible.

2. Detailed Editorial  
We need to recover a bijection A: {cells}→{1…n²} so that for each cell (i,j):  
  · left[i][j]  equals count of cells (i,k) with k<j and A[i][k]>A[i][j],  
  · top[i][j]   equals count of cells (k,j) with k<i and A[k][j]>A[i][j].  

Key observations:  
1. Feasibility check. In any row i, at column j we can have at most j larger entries to the left, so left[i][j] ≤ j. Similarly top[i][j] ≤ i. If any violation, answer is 0.  
2. Relative order constraints. For each row separately, the sequence {left[i][j]}₀≤j<n is exactly the “inversion counts” (Lehmer code) of a permutation of size n, describing the relative order of A[i][0],…,A[i][n−1] by value. Likewise for each column.  
3. Graph model. Create one node per cell. For each row i, reconstruct the permutation P of its n cells by interpreting left[i] as a Lehmer code (using a Fenwick tree or order‐statistics structure). If in that row the kth-smallest cell (by A-value) is u and the (k−1)th-smallest is v, we know A[u]>A[v]? Actually the conventional Lehmer‐code construction yields the mapping from positions to relative ranks; then to enforce the chain of increasing values we add a directed edge from the smaller-value node to the larger-value node (so in topological order smaller comes first). Repeat for columns using top[].  
4. Topological sort. We now have a DAG with O(n²) nodes and O(n²) edges. A valid topological ordering gives a labeling 1…n² in order, satisfying all row/column constraints. If the graph has a cycle (or we never reach all nodes in the sort), print 0; otherwise output the labels in matrix form.

Complexities:  
– Feasibility check: O(n²).  
– Building each row/column permutation via Fenwick: O(n log n) each, total O(n² log n).  
– DAG construction: O(n²).  
– Toposort: O(n²).  
Overall O(n² log n), n≤600 is comfortable under 0.5 s in C++ and just fits in an optimized Python.

3. Provided C++ Solution with Line-by-Line Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Overload for easy pair printing/reading
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;
}
// Overload for vector I/O
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;
}

// Fenwick tree for order-statistics (1-based)
template<class T>
class Fenwick {
  private:
    int sz;
    vector<T> tr;
  public:
    void init(int n) {
        sz = n + 1;
        tr.assign(sz + 1, 0);
    }
    // point update: add val at index idx
    void update(int idx, T val) {
        for(; idx <= sz; idx += idx & -idx)
            tr[idx] += val;
    }
    // prefix sum query
    T query(int idx) {
        T s = 0;
        for(; idx > 0; idx -= idx & -idx)
            s += tr[idx];
        return s;
    }
    // range sum
    T query(int l, int r) {
        return query(r) - query(l - 1);
    }
    // find smallest idx so that prefix sum >= k
    int find_kth(T k) {
        int idx = 0;
        // assume sz <= 2^21
        for(int bit = 1 << 20; bit > 0; bit >>= 1) {
            if(idx + bit <= sz && tr[idx + bit] < k) {
                k -= tr[idx + bit];
                idx += bit;
            }
        }
        return idx + 1;
    }
};

int n;
vector<vector<int>> up_larger, left_larger;

// Read input
void read() {
    cin >> n;
    up_larger.assign(n, vector<int>(n));
    left_larger.assign(n, vector<int>(n));
    cin >> up_larger >> left_larger;
}

// Solve one test
void solve() {
    // 1) Quick feasibility: left[i][j] ≤ j and up[i][j] ≤ i
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            if(left_larger[i][j] > j || up_larger[i][j] > i) {
                cout << 0 << "\n";
                return;
            }
        }
    }

    // Prepare adjacency list for n*n nodes
    vector<vector<int>> adj(n * n);
    Fenwick<int> fenw; 
    fenw.init(n);

    // 2) Build row constraints
    for(int i = 0; i < n; i++) {
        // Initialize Fenwick to all 1's (all positions available)
        for(int p = 1; p <= n; p++) fenw.update(p, 1);

        // vals[r] = the node-index whose rank in this row is r (0=smallest)
        vector<int> vals(n);
        // Process columns right-to-left to decode Lehmer code
        for(int j = n - 1; j >= 0; j--) {
            // we want the (left_larger[i][j] + 1)-th free spot
            int pos = fenw.find_kth(left_larger[i][j] + 1);
            int node = i * n + j;        // global node id
            vals[pos - 1] = node;        // store it by rank
            fenw.update(pos, -1);        // mark that rank as used
        }
        // Now vals[0] is smallest, vals[1] second smallest, …
        // Add edge: vals[k] → vals[k+1] for increasing values
        for(int k = 0; k + 1 < n; k++) {
            adj[vals[k]].push_back(vals[k+1]);
        }
    }

    // 3) Build column constraints similarly
    for(int j = 0; j < n; j++) {
        for(int p = 1; p <= n; p++) fenw.update(p, 1);
        vector<int> vals(n);
        for(int i = n - 1; i >= 0; i--) {
            int pos = fenw.find_kth(up_larger[i][j] + 1);
            int node = i * n + j;
            vals[pos - 1] = node;
            fenw.update(pos, -1);
        }
        for(int k = 0; k + 1 < n; k++) {
            adj[vals[k]].push_back(vals[k+1]);
        }
    }

    // 4) Topological sort
    vector<int> indeg(n * n, 0);
    for(int u = 0; u < n*n; u++)
        for(int v: adj[u])
            indeg[v]++;

    queue<int> q;
    for(int u = 0; u < n*n; u++)
        if(indeg[u] == 0)
            q.push(u);

    vector<vector<int>> A(n, vector<int>(n));
    int label = 1;
    while(!q.empty()) {
        int u = q.front(); q.pop();
        int i = u / n, j = u % n;
        A[i][j] = label++;
        for(int v: adj[u]) {
            if(--indeg[v] == 0)
                q.push(v);
        }
    }
    // If not all labeled, cycle ⇒ impossible
    if(label != n*n + 1) {
        cout << 0 << "\n";
        return;
    }

    // 5) Print the reconstructed matrix
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++)
            cout << A[i][j] << ' ';
        cout << "\n";
    }
}

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

4. Python Solution with Detailed Comments  
```python
import sys
import threading
def main():
    sys.setrecursionlimit(10**7)
    data = sys.stdin.read().split()
    it = iter(data)
    try:
        n = int(next(it))
    except StopIteration:
        return
    # Read matrices
    up    = [[int(next(it)) for _ in range(n)] for _ in range(n)]
    left  = [[int(next(it)) for _ in range(n)] for _ in range(n)]
    # 1) Feasibility check
    for i in range(n):
        for j in range(n):
            if left[i][j] > j or up[i][j] > i:
                print(0)
                return

    # Fenwick tree supporting point updates and prefix sums
    class Fenw:
        def __init__(self, n):
            self.n = n
            self.f = [0]*(n+1)
        def update(self, i, v):
            # add v at index i (1-based)
            while i <= self.n:
                self.f[i] += v
                i += i & -i
        def query(self, i):
            # sum f[1..i]
            s = 0
            while i>0:
                s += self.f[i]
                i -= i & -i
            return s
        def find_kth(self, k):
            # find smallest i with prefix sum ≥ k
            idx = 0
            bit = 1<<(self.n.bit_length())
            while bit>0:
                nxt = idx+bit
                if nxt<=self.n and self.f[nxt]<k:
                    k -= self.f[nxt]
                    idx = nxt
                bit >>= 1
            return idx+1

    # Build adjacency list for DAG of size n*n
    N = n*n
    adj = [[] for _ in range(N)]
    indeg = [0]*N

    # Helper to add chain edges given a Lehmer code row/col
    def process_block(code, is_row, idx):
        # code: list of length n of inversion counts
        # is_row: True if it's a row, idx is row-index, else column
        fenw = Fenw(n)
        for i in range(1, n+1):
            fenw.update(i, 1)
        vals = [0]*n
        # decode from right to left
        for pos in range(n-1, -1, -1):
            cnt = code[pos]
            place = fenw.find_kth(cnt+1)
            node = (idx*n + pos) if is_row else (pos*n + idx)
            vals[place-1] = node
            fenw.update(place, -1)
        # now vals[0]..vals[n-1] are nodes in ascending A-value order
        for k in range(n-1):
            u = vals[k]
            v = vals[k+1]
            adj[u].append(v)
            indeg[v] += 1

    # 2) rows
    for i in range(n):
        process_block(left[i], True, i)
    # 3) columns
    for j in range(n):
        col_code = [up[i][j] for i in range(n)]
        process_block(col_code, False, j)

    # 4) Topological sort (Kahn)
    from collections import deque
    dq = deque(u for u in range(N) if indeg[u]==0)
    ans = [0]*N
    cur = 1
    while dq:
        u = dq.popleft()
        ans[u] = cur
        cur += 1
        for v in adj[u]:
            indeg[v] -= 1
            if indeg[v]==0:
                dq.append(v)
    if cur != N+1:
        print(0)
        return

    # 5) Print matrix
    out = []
    for i in range(n):
        row = ans[i*n:(i+1)*n]
        out.append(" ".join(map(str,row)))
    print("\n".join(out))

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

5. Compressed Editorial  
- Check left[i][j] ≤ j and top[i][j] ≤ i or output 0.  
- For each row/column, interpret the counts as a Lehmer code, rebuild the relative order of its n cells via an order-statistic (Fenwick) tree, producing a sorted list of node ids by value.  
- Add edges from each k‐th to (k+1)‐th in that list, enforcing increasing A-values.  
- Topologically sort the combined DAG of n² nodes.  
  · If cycle → print 0.  
  · Else assign labels 1…n² in topo order and output the matrix.  
Complexity O(n² log n).