1. Abridged Problem Statement  
Given an N×M integer matrix A, construct an N×M matrix B such that for each 0≤i<N and 0≤j<M,  
 B[i][j] = min { A[x][y] : y ≥ j and x ≥ i + j – y }.

2. Detailed Editorial  
Goal  
----  
For each cell (i,j) in A, we want the minimum over all A[x][y] satisfying two linear inequalities:  
 1) y ≥ j  
 2) x ≥ i + j – y  

Rewriting the second condition:  
 x + y ≥ i + j.  

So the feasible (x,y) form a “down-and-right” region when viewed in a transformed coordinate system. We need to answer one rectangular‐range minimum per cell efficiently (N,M up to 1,000, so an O(N·M) or O(N·M·log) approach is required).

Key Observation: Diagonal Index  
-------------------------------  
Define s = x + y. Then the conditions become:  
 y ≥ j  
 s ≥ i + j  

Thus all points (x,y) that contribute to B[i][j] lie in the set  
 { (x,y) | y = row-index from j..M−1, s = x+y from (i+j)..(N−1 + M−1) }.

If we build a 2D array Q of size M rows (indexed by y) and N+M−1 columns (indexed by s), and initialize  
 Q[y][s] = A[s−y][y] whenever 0 ≤ s−y < N,  
else Q[y][s] = +∞,  
then the answer B[i][j] sits at Q[j][i+j] after we precompute in Q the minimum over the suffix region (down & right) at each cell.

Dynamic Programming on Q  
-------------------------  
We scan Q in reverse order of rows and columns (from bottom‐right to top‐left). For each Q[i][j], we set:  
 Q[i][j] = min( Q[i][j], Q[i+1][j], Q[i][j+1] )  
where out‐of‐bounds indices yield +∞. This ensures Q[i][j] stores the minimum over its own cell and all cells reachable by going “down” or “right” in Q, exactly matching our desired region for B.

Finally, map back:  
 B[i][j] = Q[j][i+j].

Complexity  
----------  
Time: O((N+M)·M + N·M) = O(N·M).  
Memory: O(M·(N+M)) ~ up to 2×10^6 entries, stored as 16-bit or 32-bit numbers.

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;
};

int n, m;
vector<vector<short>> a;

void read() {
    cin >> n >> m;
    a.assign(n, vector<short>(m));
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            cin >> a[i][j];
        }
    }
}

void solve() {
    // B[i][j] = min{ A[x][y] : y >= j and x >= i + j - y }, i.e. the minimum
    // over the lower-left quadrant relative to the anti-diagonal through
    // (i, j). Reindex by column y and anti-diagonal d = x + y: cell A[x][y]
    // lands at q[y][x + y]. The constraints y >= j and x + y >= i + j become
    // "row index >= j and diagonal index >= i + j", so each B value is a
    // suffix minimum of q over both indices. We compute that suffix minimum
    // with a backward sweep (each cell takes the min of itself, the cell
    // below and the cell to the right), then read B[i][j] back from
    // q[j][i + j].

    const short INF = numeric_limits<short>::max();
    vector<vector<short>> q(m, vector<short>(n + m, INF));

    for(int x = 0; x < n; x++) {
        for(int y = 0; y < m; y++) {
            q[y][x + y] = a[x][y];
        }
    }

    for(int i = m - 1; i >= 0; i--) {
        for(int j = n + m - 1; j >= 0; j--) {
            short cur = q[i][j];
            short down = (i + 1 < m) ? q[i + 1][j] : INF;
            short right = (j + 1 < n + m) ? q[i][j + 1] : INF;
            q[i][j] = min({cur, down, right});
        }
    }

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            cout << q[j][i + j] << (j == m - 1 ? '\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

def solve(N, M, A):
    INF = 10**9
    # Build Q: M rows, N+M columns
    Q = [[INF] * (N + M) for _ in range(M)]

    # Map A into Q by diagonal index s = x+y
    for x in range(N):
        for y in range(M):
            s = x + y
            Q[y][s] = A[x][y]

    # DP: compute min over suffix region (down and right moves)
    # iterate rows and cols in reverse
    for i in range(M - 1, -1, -1):
        for j in range(N + M - 1, -1, -1):
            curr = Q[i][j]
            # candidate from cell below
            down = Q[i + 1][j] if i + 1 < M else INF
            # candidate from cell to the right
            right = Q[i][j + 1] if j + 1 < N + M else INF
            # store the minimum of the three
            Q[i][j] = min(curr, down, right)

    # Extract B by reversing the mapping
    B = [[0]*M for _ in range(N)]
    for i in range(N):
        for j in range(M):
            s = i + j
            B[i][j] = Q[j][s]
    return B

def main():
    data = sys.stdin.read().split()
    it = iter(data)
    N, M = map(int, (next(it), next(it)))
    A = [[int(next(it)) for _ in range(M)] for _ in range(N)]
    B = solve(N, M, A)
    out = []
    for row in B:
        out.append(" ".join(map(str, row)))
    sys.stdout.write("\n".join(out))

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

5. Compressed Editorial  
- Define s = x+y, y = column index.  
- Build Q[y][s] = A[x][y] or +∞ if outside.  
- Run DP: Q[i][j] = min(Q[i][j], Q[i+1][j], Q[i][j+1]) in reverse order.  
- Then B[i][j] = Q[j][i+j].  
- Overall O(N·M) time, O(M·(N+M)) space.