<|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.

354. Just Matrix
Time limit per test: 0.5 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



This task is not about Berland or Beerland troubles, roads or flights. There will be no boring coins. This task is about simple square matrix.

Numbers from 1 to n2 were written down in nx n square matrix A. Each number was written exactly once. After that for each number the pair topi,j and lefti,j was written. topi,j is the number of elements in column j bigger than Ai,j and positioned to the top from Ai,j. lefti,j is the number of elements in the row i bigger than Ai,j and positioned to the left from Ai,j.

You are given matrices top and left. Your task is to find any possible matrix A fulfilling the requirements of the problem.

Input
The first line of the input contains integer number n (1 ≤ n ≤ 600). Further matrices top and left are written, each in the form of n lines of n non-negative integer numbers. The matrices are separated by the empty line. Numbers in both matrices are not bigger than n.

Output
Write to the output matrix A in the format similar to the input data. If there are several solutions, you can choose any of them. If there is no solution, write to the output just one number 0.

Example(s)
sample input
sample output
3
0 0 0 
0 0 0
0 0 2

0 0 0
0 1 0
0 1 2
1 2 6
5 3 7
9 8 4

<|response|>
1. Abridged problem statement  
You are given an integer n and two n×n non‐negative integer matrices top and left. There was an unknown permutation of numbers 1…n² placed into an n×n matrix A so that for each cell (i,j):  
  • left[i][j] = the count of entries in row i to the left of (i,j) that are larger than A[i][j],  
  • top[i][j]  = the count of entries in column j above (i,j) that are larger than A[i][j].  
Reconstruct any valid A or report 0 if no such A exists.

2. Key observations  
- Feasibility check: In row i, position j has at most j cells to its left, so left[i][j] ≤ j.  Similarly, top[i][j] ≤ i.  If violated, answer is 0.  
- A row’s sequence of “left counts” is exactly the Lehmer code (inversion vector) of the permutation of that row’s values.  Likewise each column’s “top counts” is a Lehmer code.  
- From a Lehmer code of length n you can reconstruct the relative ordering of n elements by maintaining a multiset of free ranks {1…n}, repeatedly taking the (code[pos]+1)-th smallest unused rank.  That yields an ordering from smallest value to largest.  
- Enforce these orderings as a directed chain: if in some row the k-th smallest cell (by value) is u and the (k+1)-th smallest is v, add edge u → v.  Do the same for each column.  
- The union of all such edges is a DAG on n² nodes.  A topological ordering of that DAG assigns distinct labels 1…n² consistent with all row/column constraints.  If the graph has a cycle, print 0.

3. Full solution approach  
a) Read n, the matrices top and left.  
b) Check for any (i,j) that left[i][j] > j or top[i][j] > i.  If so, print 0 and exit.  
c) Number each cell by a unique node id = i·n + j (0 ≤ id < n²).  Prepare adjacency list adj of size n².  
d) Define a Fenwick‐tree (BIT) structure over [1…n] supporting  
   - point updates,  
   - prefix‐sum queries,  
   - find_kth(k): the smallest index p with prefix_sum(p) ≥ k.  
e) For each row i:  
   • Initialize Fenwick so every index 1…n has value 1 (n free ranks).  
   • Let code[j] = left[i][j].  Process j from n−1 down to 0:  
       – want the (code[j]+1)-th free rank → p = find_kth(code[j]+1).  
       – set vals[p−1] = node_id(i,j).  
       – update Fenwick at p by −1 (mark rank used).  
   • Now vals[0…n−1] lists the row’s nodes in increasing A-value order.  Chain consecutive cells with a directed edge (the implementation orients it from vals[k+1] to vals[k]); any consistent orientation that makes the toposort respect the order works.
f) For each column j: do the same using code[i] = top[i][j], mapping node_id(i,j), and add chain edges.  
g) Compute indegree[] for all nodes, then run Kahn’s algorithm:  
   • enqueue all nodes of indegree 0, repeatedly pop u, assign A[i][j] = next_label++ to cell u, decrement indegree of its neighbors, enqueue any that drop to 0.  
h) If you assigned fewer than n² labels (cycle detected), print 0; otherwise print the matrix A.

Complexity: O(n² log n) for decoding all rows/columns + O(n²) for topo sort.

4. C++ implementation
```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;
}
```

5. Python implementation with detailed comments  
```python
import sys
from collections import deque

def main():
    data = sys.stdin.read().split()
    if not data: 
        return
    it = iter(data)
    n = int(next(it))

    # Read top and left matrices
    top = [ [int(next(it)) for _ in range(n)] for _ in range(n) ]
    leftc = [ [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 leftc[i][j] > j or top[i][j] > i:
                print(0)
                return

    # Prepare adjacency for n^2 nodes
    N = n * n
    adj = [[] for _ in range(N)]
    indeg = [0] * N

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

    fenw = Fenw(n)

    # map (i,j) to node id
    def node_id(i,j):
        return i * n + j

    # decode a Lehmer code array of length n
    # getNode(pos) returns the node_id for that position
    def decode_and_chain(code, getNode):
        # reset Fenwick: mark all ranks [1..n] free
        for x in range(1, n+1):
            fenw.update(x, 1)
        vals = [0]*n
        # assign from rightmost position down to 0
        for pos in range(n-1, -1, -1):
            cnt = code[pos]
            rank = fenw.find_kth(cnt + 1)
            vals[rank - 1] = getNode(pos)
            fenw.update(rank, -1)
        # add chain edges
        for k in range(n-1):
            u = vals[k]
            v = vals[k+1]
            adj[u].append(v)
            indeg[v] += 1

    # 2) process rows
    for i in range(n):
        decode_and_chain(leftc[i], lambda pos, i=i: node_id(i, pos))

    # 3) process columns
    for j in range(n):
        col_code = [ top[i][j] for i in range(n) ]
        decode_and_chain(col_code, lambda pos, j=j: node_id(pos, j))

    # 4) Kahn's algorithm for topo sort
    q = deque(u for u in range(N) if indeg[u] == 0)
    Aflat = [0] * N
    label = 1
    while q:
        u = q.popleft()
        Aflat[u] = label
        label += 1
        for v in adj[u]:
            indeg[v] -= 1
            if indeg[v] == 0:
                q.append(v)

    # if cycle detected
    if label != N+1:
        print(0)
        return

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


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