1. Abridged Problem Statement  
Given an N×M grid where some cells initially contain weed (‘X’) and others are empty (‘.’), weed spreads over the years so that any empty cell with at least two weed‐occupied neighbors (up/down/left/right) eventually becomes occupied. Compute the total number of weed‐occupied cells after the spread stabilizes.

2. Detailed Editorial  
We need to model the spread process until no more cells turn from empty to weed. A naive simulation that repeatedly scans the whole grid and applies the rule in rounds would be O((NM)^2) in the worst case, too big for N,M up to 1000. Instead, observe that once a cell becomes weed, it contributes to its neighbors’ “weed‐count”; an empty cell turns to weed exactly when its count of weed neighbors reaches two. We can implement this with a one‐pass BFS-like propagation:

• Data Structures  
  • visited[i][j] (or a flat array) marks cells that have become weed, including the initial ones.  
  • degree[i][j] counts how many of its four neighbors are already marked weed.  
  • queue Q holds cells that have just become weed and whose neighbors we must update.

• Initialization  
  – Push all initial weed cells into Q and mark them visited.  
  – degree[*] = 0.

• BFS Propagation  
  While Q is not empty:  
    – Pop cell u.  
    – For each of its four neighbors v inside the grid:  
        • Increment degree[v].  
        • If degree[v] reaches 2 and v is not yet visited:  
            – Mark v visited, push v into Q.

Every cell is enqueued at most once, and every edge (adjacency) is examined once—O(NM) time. Finally, the answer is the total number of visited cells.

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

// Overload for printing a pair<T1,T2>
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &x) {
    return out << x.first << ' ' << x.second;
}

// Overload for reading a pair<T1,T2>
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &x) {
    return in >> x.first >> x.second;
}

// Overload for reading a vector<T>
template<typename T>
istream &operator>>(istream &in, vector<T> &a) {
    for(auto &x: a) {
        in >> x;
    }
    return in;
}

// Overload for printing a vector<T>
template<typename T>
ostream &operator<<(ostream &out, const vector<T> &a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
}

int n, m;                  // Dimensions of the grid
vector<string> t;          // The grid itself, as strings of 'X' and '.'

// Read input: n, m and then n rows of the grid
void read() {
    cin >> n >> m;
    t.resize(n);
    cin >> t;
}

void solve() {
    int k = n * m;                      // Total number of cells
    vector<bool> visited(k, false);     // visited[u] = true if cell u has weed
    vector<int> degree(k, 0);           // degree[u] = count of weed neighbors

    queue<int> q;                       // BFS queue storing linearized cell indices

    // 1) Initialize: push all initially weed-occupied cells
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            if(t[i][j] == 'X') {
                int u = i * m + j;      // linear index
                q.push(u);
                visited[u] = true;
            }
        }
    }

    int ans = 0;                        // counter for total weed cells

    // 2) BFS propagation
    while(!q.empty()) {
        int u = q.front(); q.pop();
        ans++;                          // this cell is confirmed weed
        int i = u / m;                  // row index
        int j = u % m;                  // column index

        // Explore four neighbors
        static const int di[4] = {-1,1,0,0};
        static const int dj[4] = {0,0,-1,1};
        for(int dir = 0; dir < 4; dir++) {
            int ni = i + di[dir];
            int nj = j + dj[dir];
            // Check bounds
            if(ni >= 0 && ni < n && nj >= 0 && nj < m) {
                int v = ni * m + nj;
                degree[v]++;            // one more weed neighbor
                // When a non-visited cell accumulates two weed neighbors, it turns weed
                if(degree[v] >= 2 && !visited[v]) {
                    visited[v] = true;
                    q.push(v);
                }
            }
        }
    }

    // Output the total number of weed cells after spread
    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
from collections import deque

def main():
    input = sys.stdin.readline
    n, m = map(int, input().split())
    grid = [input().rstrip('\n') for _ in range(n)]

    # Flattened indexing: u = i*m + j
    k = n * m
    visited = [False] * k       # True if this cell currently has weed
    degree = [0] * k            # Number of weed neighbors

    q = deque()
    # Enqueue initial weed cells
    for i in range(n):
        for j in range(m):
            if grid[i][j] == 'X':
                u = i * m + j
                visited[u] = True
                q.append(u)

    ans = 0
    # Directions: up, down, left, right
    dirs = [(-1,0),(1,0),(0,-1),(0,1)]

    # BFS-like propagation
    while q:
        u = q.popleft()
        ans += 1
        i, j = divmod(u, m)
        # Update each neighbor
        for di, dj in dirs:
            ni, nj = i + di, j + dj
            if 0 <= ni < n and 0 <= nj < m:
                v = ni * m + nj
                degree[v] += 1
                # If an empty cell now has ≥2 weed neighbors, it becomes weed
                if degree[v] == 2 and not visited[v]:
                    visited[v] = True
                    q.append(v)

    print(ans)

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

5. Compressed Editorial  
Perform a single BFS-style propagation: start from all ‘X’ cells, maintain for each empty cell the tally of weed neighbors, and the moment it reaches two, enqueue it as newly weed. This visits each cell once and each adjacency once, yielding O(NM) time.