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 relative value order of its n cells by interpreting left[i] as a Lehmer code (using a Fenwick tree / order‐statistics structure with find_kth): decoding right-to-left gives, for each rank, which cell holds it, so we obtain the cells of the row sorted by A-value. Consecutive cells in this sorted list are then chained with directed edges so that the topological order respects their increasing values. Repeat for columns using top[]. (The edges connect each cell to its neighbor of adjacent value; the precise orientation only has to be consistent, which it is.)
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. C++ Solution
```cpp
#include <bits/stdc++.h>
// #include <coding_library/data_structures/fenwick.hpp>

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;
};

template<class T>
class Fenwick {
  private:
    int sz, log_size;
    vector<T> tr;

  public:
    void init(int n) {
        sz = n + 1;
        log_size = 31 - __builtin_clz(sz);
        tr.assign(sz + 1, 0);
    }

    void update(int idx, T val) {
        if(idx <= 0) {
            assert(false);
            return;
        }
        for(; idx <= sz; idx += (idx & -idx)) {
            tr[idx] += val;
        }
    }

    T query(int idx) {
        T ans = 0;
        for(; idx >= 1; idx -= (idx & -idx)) {
            ans += tr[idx];
        }

        return ans;
    }

    T query(int l, int r) { return query(r) - query(l - 1); }

    int find_kth(T k) {
        int idx = 0;
        for(int i = log_size; i >= 0; i--) {
            if(idx + (1 << i) < sz && tr[idx + (1 << i)] < k) {
                k -= tr[idx + (1 << i)];
                idx += (1 << i);
            }
        }
        return idx + 1;
    }
};

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

void read() {
    cin >> n;
    up_larger.assign(n, vector<int>(n, 0));
    left_larger.assign(n, vector<int>(n, 0));
    cin >> up_larger >> left_larger;
}

void solve() {
    // We can solve this problem by creating a compressed graph with O(n^2)
    // nodes and O(n^2) edges and then finding a topological sort of it. In
    // particular,  for every row and column, we can solve the more-standard
    // problem in a permutation, find the orders of the cells, and then add the
    // corresponding n-1 edges between consecutive cells. Finding the order in
    // every row and column can be done in O(n log n) time, and then we can find
    // the topological sort in O(n^2) time, making the total complexity O(n^2
    // log n).
    // We can notice that the answer is 0 if and only if we can't construct the
    // order in a row / column, or if there is a cycle in the graph. The former
    // happens only when up_larger[i][j] > i or left_larger[i][j] > j, as
    // otherwise we always have a construction.

    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;
            }
        }
    }

    Fenwick<int> fenwick;
    fenwick.init(n + 4);

    vector<vector<int>> adj(n * n);
    for(int i = 0; i < n; i++) {
        for(int j = 1; j <= n; j++) {
            fenwick.update(j, 1);
        }

        vector<int> vals(n, 0);
        for(int j = n - 1; j >= 0; j--) {
            int pos = fenwick.find_kth(left_larger[i][j] + 1);
            int idx = i * n + j;
            vals[pos - 1] = idx;
            fenwick.update(pos, -1);
        }

        for(int j = 1; j < n; j++) {
            adj[vals[j]].push_back(vals[j - 1]);
        }
    }

    for(int j = 0; j < n; j++) {
        for(int i = 1; i <= n; i++) {
            fenwick.update(i, 1);
        }

        vector<int> vals(n, 0);
        for(int i = n - 1; i >= 0; i--) {
            int pos = fenwick.find_kth(up_larger[i][j] + 1);
            int idx = i * n + j;
            vals[pos - 1] = idx;
            fenwick.update(pos, -1);
        }

        for(int i = 1; i < n; i++) {
            adj[vals[i]].push_back(vals[i - 1]);
        }
    }

    vector<vector<int>> ans(n, vector<int>(n, 0));

    queue<int> q;
    int current_node = 1;
    vector<int> in_degree(n * n, 0);

    for(int u = 0; u < n * n; u++) {
        for(int v: adj[u]) {
            in_degree[v]++;
        }
    }

    for(int u = 0; u < n * n; u++) {
        if(in_degree[u] == 0) {
            q.push(u);
        }
    }

    while(!q.empty()) {
        int u = q.front();
        q.pop();
        int i = u / n, j = u % n;
        ans[i][j] = current_node++;

        for(int v: adj[u]) {
            in_degree[v]--;
            if(in_degree[v] == 0) {
                q.push(v);
            }
        }
    }

    if(current_node != n * n + 1) {
        cout << 0 << '\n';
        return;
    }

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            cout << ans[i][j] << ' ';
        }
        cout << '\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():
    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).