## 1) Concise, abridged problem statement

You are given an `N × M` grid (`1 ≤ N, M ≤ 180`) of `0/1` where `1` is a tree (blocked) and `0` is empty.

You must place a “C”-shaped house composed of **three non-overlapping rectangles** (top, middle, bottom) such that:

1. Each rectangle has **positive area**.
2. Rectangles **touch** each other (adjacent along edges), but **do not overlap**.
3. The **top-left cell** of all three rectangles lies on the **same column** (same left edge).
4. The **middle rectangle’s width is strictly smaller** than the widths of both top and bottom rectangles.
5. The house can only occupy `0` cells.

Maximize total area.  
Output the maximum area and the grid with house cells marked as `8` (trees remain `1`, other empty cells remain `0`). If impossible, output `-1`.

---

## 2) Detailed editorial (explaining the provided solution)

### Key observation: Fix the left edge (column)
All three rectangles must start at the same left column `c`. If we fix `c`, then for each row `r` we can compute:

- `dist[r][c]` = how many consecutive empty cells (`0`) exist starting at `(r, c)` going right.

This gives us a “histogram” of possible widths for rectangles whose left edge is at column `c`:
`height[r] = dist[r][c]`.

Any rectangle that starts at column `c` and spans some consecutive rows `[a..b]` can have width at most:
`min(height[a..b])`.

This is the same setup as the classic “largest rectangle in a histogram / maximal rectangle in binary matrix” approach.

### Precompute `dist`
For each row, compute distances to the next tree to the right:

- If `grid[row][col] == 1` (tree), `dist=0`.
- Else `dist[row][col] = 1 + dist[row][col+1]` (or 1 at last column).

This is `O(N*M)`.

### For a fixed column `c`, solve best “C” in `O(N^2)`
We need 3 rectangles stacked vertically: top, middle, bottom, all starting at column `c`.

Let:
- top uses some rows ending above middle,
- middle uses some rows between,
- bottom uses rows below.

Widths:
- `w1` (top), `w2` (middle), `w3` (bottom),
- must satisfy `w2 < w1` and `w2 < w3` and all > 0.

The solution iterates over:
- a “critical” row `i` that determines the top width `w1 = height[i]`,
- a “critical” row `j` (with `j >= i+2`) that determines the bottom width `w3 = height[j]`.

Then it tries to place the middle section between them.

#### Nearest smaller elements: `l[]` and `r[]`
For each row `k`, compute:
- `l[k]` = nearest index above `k` with `height[l[k]] < height[k]` (or `-1`)
- `r[k]` = nearest index below `k` with `height[r[k]] < height[k]` (or `n`)

These are computed with monotonic stacks in `O(N)`.

Interpretation:
- The maximal vertical span where `height >= height[k]` is `(l[k]+1 .. r[k]-1)`.

So:
- Top rectangle with width `w1=height[i]` can extend up/down within those bounds, but in our construction we only use it above the middle.
- Bottom rectangle with width `w3=height[j]` can extend similarly.

#### Middle section cases
We want the middle width `w2` to be as large as possible but strictly smaller than both `w1` and `w3`.

The code handles three configurations depending on overlap of the maximal spans:

**Case 1: `r[i] >= j`**  
The “tall-enough” region for `i` reaches down to (or beyond) `j`.  
So between top and bottom there is no guaranteed gap; we can choose a **single middle row** placed near the boundary.

- Pick a middle row `mid_row` near `j`’s left boundary: `mid_row = max(i+1, l[j])` (as in code)
- Set `w2 = min(w1, w3) - 1` (strictly smaller than both)
- Compute area:
  - top part rows `[top_start..top_end]` width `w1`
  - middle part 1 row width `w2`
  - bottom part rows `[bot_start..bot_end]` width `w3`

Check `w2 > 0` (required), maximize area.

**Case 2: `l[j] <= i`**  
Symmetric to case 1, place a 1-row middle close to `i`’s right boundary.

**Case 3: `r[i] <= l[j]`**  
Now there is a real “middle band” of rows between them. The middle rectangle can have height > 1.

But its width is limited by:
- it must be < `w1` and < `w3`, i.e. `w2 ≤ w1-1` and `w2 ≤ w3-1`
- and must fit all rows in the middle band: `w2 ≤ min(height[mid_start..mid_end])`

To answer `min(height[L..R])` quickly, the code precomputes a quadratic table:

`min_h[a][b] = min(height[a..b])` in `O(N^2)`.

Then:
- `min_middle = min_h[r[i]][l[j]]`
- `w2 = min(w1-1, w3-1, min_middle)`

Area is computed as three rectangles: top band, middle band, bottom band.

#### Complexity
For each column:
- build `l,r` in `O(N)`
- build `min_h` in `O(N^2)`
- loop over `(i,j)` in `O(N^2)` with O(1) case evaluation

So per column: `O(N^2)`. Total: `O(M*N^2)`.

With `N,M ≤ 180`, worst case ~ `180*180^2 ≈ 5.8M` iterations plus small constants, fits in 0.25s in optimized C++ (tight but plausible with low overhead).

### Reconstruction / output
The solver stores the best configuration: column, row ranges for the three parts, and widths.
Then it writes `8` into those cells in the original `grid` and prints.

If no valid “C” found, print `-1`.

---

## 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<int>> grid;
vector<vector<int>> dist;

struct Result {
    int area;
    int col;
    int r1_start, r1_end, w1;
    int r2_start, r2_end, w2;
    int r3_start, r3_end, w3;

    Result()
        : area(-1),
          col(-1),
          r1_start(-1),
          r1_end(-1),
          w1(-1),
          r2_start(-1),
          r2_end(-1),
          w2(-1),
          r3_start(-1),
          r3_end(-1),
          w3(-1) {}

    Result(
        int a, int c, int rs1, int re1, int ww1, int rs2, int re2, int ww2,
        int rs3, int re3, int ww3
    )
        : area(a),
          col(c),
          r1_start(rs1),
          r1_end(re1),
          w1(ww1),
          r2_start(rs2),
          r2_end(re2),
          w2(ww2),
          r3_start(rs3),
          r3_end(re3),
          w3(ww3) {}
};

void read() {
    cin >> n >> m;
    grid.assign(n, vector<int>(m));
    cin >> grid;
    dist.assign(n, vector<int>(m));
    for(int row = 0; row < n; row++) {
        for(int col = m - 1; col >= 0; col--) {
            if(grid[row][col] == 1) {
                dist[row][col] = 0;
            } else {
                dist[row][col] = (col == m - 1) ? 1 : dist[row][col + 1] + 1;
            }
        }
    }
}

Result solve_histogram(int col) {
    vector<int> height(n);
    for(int row = 0; row < n; row++) {
        height[row] = dist[row][col];
    }

    vector<int> l(n), r(n);
    stack<int> s;

    for(int i = 0; i < n; i++) {
        while(!s.empty() && height[s.top()] >= height[i]) {
            s.pop();
        }
        l[i] = s.empty() ? -1 : s.top();
        s.push(i);
    }
    while(!s.empty()) {
        s.pop();
    }

    for(int i = n - 1; i >= 0; i--) {
        while(!s.empty() && height[s.top()] >= height[i]) {
            s.pop();
        }
        r[i] = s.empty() ? n : s.top();
        s.push(i);
    }

    vector<vector<int>> min_h(n, vector<int>(n, INT_MAX));
    for(int i = 0; i < n; i++) {
        min_h[i][i] = height[i];
        for(int j = i + 1; j < n; j++) {
            min_h[i][j] = min(min_h[i][j - 1], height[j]);
        }
    }

    Result best;
    for(int i = 0; i < n; i++) {
        if(height[i] == 0) {
            continue;
        }

        int top_start = l[i] + 1, top_end;
        for(int j = i + 2; j < n; j++) {
            if(height[j] == 0) {
                continue;
            }

            int bot_start;
            int bot_end = r[j] - 1;
            int w1 = height[i];
            int w3 = height[j];

            if(r[i] >= j) {
                int mid_row = max(i + 1, l[j]);
                top_end = mid_row - 1;
                bot_start = mid_row + 1;

                int w2 = min(w1, w3) - 1;
                int area = (top_end - top_start + 1) * w1 + 1 * w2 +
                           (bot_end - bot_start + 1) * w3;
                if(w2 > 0 && area > best.area) {
                    best = Result(
                        area, col, top_start, top_end, w1, mid_row, mid_row, w2,
                        bot_start, bot_end, w3
                    );
                }
            }

            if(l[j] <= i) {
                int mid_row = min(j - 1, r[i]);
                top_end = mid_row - 1;
                bot_start = mid_row + 1;

                int w2 = min(w1, w3) - 1;
                int area = (top_end - top_start + 1) * w1 + 1 * w2 +
                           (bot_end - bot_start + 1) * w3;
                if(w2 > 0 && area > best.area) {
                    best = Result(
                        area, col, top_start, top_end, w1, mid_row, mid_row, w2,
                        bot_start, bot_end, w3
                    );
                }
            }

            if(r[i] <= l[j]) {
                top_end = r[i] - 1;
                bot_start = l[j] + 1;

                int mid_start = r[i];
                int mid_end = l[j];

                int min_middle = min_h[r[i]][l[j]];
                int w2 = min({w1 - 1, w3 - 1, min_middle});
                int area = (top_end - top_start + 1) * w1 +
                           (mid_end - mid_start + 1) * w2 +
                           (bot_end - bot_start + 1) * w3;
                if(w2 > 0 && area > best.area) {
                    best = Result(
                        area, col, top_start, top_end, w1, mid_start, mid_end,
                        w2, bot_start, bot_end, w3
                    );
                }
            }
        }
    }

    return best;
}

void solve() {
    // Implementing this problem is non-trivial, but the core idea is to think
    // about how the largest sub-rectangle problem is solved, and adapt the
    // solution. More precisely, there is a popular approach with building a
    // "histogram" of the heights of empty places, and then using stacks find
    // the closest columns that are smaller as they define how far we can extend
    // the current column. There is a Geeks-for-Geeks page describing the idea
    // further:
    //
    //     www.geeksforgeeks.org/dsa/maximum-size-rectangle-binary-sub-matrix-1s
    //
    // In this problem we have to choose 3 adjacent rectangles. We know that
    // their leftmost points should fall on the same vertical line. Let's try to
    // independently solve (find best possible answer) for every column in O(M).
    // We can build a histogram with the distances to the right from each cell
    // in this column. Let's try to fix i + 1 < j - the "critical" column in the
    // top and bottom rectangles out of the three. We can easily precompute with
    // a stack l[i] and r[i] as the closest positions smaller than height[i],
    // and we should trivially extend the top and bottom rectangles to l[i] and
    // r[j]. The area between i and j is a bit trickier as we have a few cases:
    //
    //     1) r[i] >= j. In that case the height of the middle column should be
    //        simply min(height[i], height[j])-1, and we can have a rectangle of
    //        length 1 directly adjacent to the larger of i and j. To be more
    //        precise, we should first try to extend j as far as possible to the
    //        left, then put the middle with width 1, and finally have the rest
    //        for i.
    //
    //     2) l[j] <= i. This is symmetric to the (1) case.
    //
    //     3) If it's neither (1) or (2), we have r[i] <= l[j]. In this case, we
    //        are bounded by min(height[k] for r[i] <= k <= l[j]) for the middle
    //        part. We can do a quadratic precompute of min[L][R] with a simple
    //        DP or a sparse table, so that this case can also be handled in
    //        constant time.
    //
    // As we already mentioned, we are looking at the O(M) histogram, finding
    // the best "C" for a histogram can be done in O(N^2) time. As we have to
    // recover the answer, it's convenient to return the relevant intervals for
    // the 3 portions, as well as the heights and the overall area. Some
    // additional parts to be careful about are heights of 0 as this won't be a
    // valid rectangle.

    Result best;
    for(int col = 0; col < m; col++) {
        Result res = solve_histogram(col);
        if(res.area > best.area) {
            best = res;
        }
    }

    if(best.area == -1) {
        cout << -1 << '\n';
        return;
    }

    cout << best.area << '\n';

    int col = best.col;
    for(int row = best.r1_start; row <= best.r1_end; row++) {
        for(int c = col; c < col + best.w1; c++) {
            grid[row][c] = 8;
        }
    }
    for(int row = best.r2_start; row <= best.r2_end; row++) {
        for(int c = col; c < col + best.w2; c++) {
            grid[row][c] = 8;
        }
    }
    for(int row = best.r3_start; row <= best.r3_end; row++) {
        for(int c = col; c < col + best.w3; c++) {
            grid[row][c] = 8;
        }
    }

    for(int row = 0; row < n; row++) {
        for(int c = 0; c < m; c++) {
            if(c > 0) {
                cout << ' ';
            }
            cout << grid[row][c];
        }
        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 (same approach) with detailed comments

```python
import sys

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

    # Read grid
    grid = [[int(next(it)) for _ in range(m)] for _ in range(n)]

    # dist[r][c] = number of consecutive empty cells (0s) from (r,c) to the right
    dist = [[0] * m for _ in range(n)]
    for r in range(n):
        # Compute from right to left
        for c in range(m - 1, -1, -1):
            if grid[r][c] == 1:
                dist[r][c] = 0
            else:
                dist[r][c] = 1 if c == m - 1 else dist[r][c + 1] + 1

    # Store best solution info
    best_area = -1
    best = None  # will hold (col, r1s,r1e,w1, r2s,r2e,w2, r3s,r3e,w3)

    # Solve independently for each left column
    for col in range(m):
        height = [dist[r][col] for r in range(n)]

        # Compute nearest strictly smaller on the left (upwards): l[i]
        l = [-1] * n
        stack = []
        for i in range(n):
            while stack and height[stack[-1]] >= height[i]:
                stack.pop()
            l[i] = stack[-1] if stack else -1
            stack.append(i)

        # Compute nearest strictly smaller on the right (downwards): r[i]
        rgt = [n] * n
        stack.clear()
        for i in range(n - 1, -1, -1):
            while stack and height[stack[-1]] >= height[i]:
                stack.pop()
            rgt[i] = stack[-1] if stack else n
            stack.append(i)

        # Precompute min_h[a][b] = min(height[a..b]) in O(n^2)
        # Using Python lists; n <= 180 so fine.
        min_h = [[10**9] * n for _ in range(n)]
        for i in range(n):
            cur = height[i]
            min_h[i][i] = cur
            for j in range(i + 1, n):
                if height[j] < cur:
                    cur = height[j]
                min_h[i][j] = cur

        # Enumerate critical rows i (top) and j (bottom)
        for i in range(n):
            if height[i] == 0:
                continue

            top_start = l[i] + 1
            w1 = height[i]

            for j in range(i + 2, n):
                if height[j] == 0:
                    continue

                w3 = height[j]
                bot_end = rgt[j] - 1

                # --- Case 1: r[i] >= j (place 1-row middle) ---
                if rgt[i] >= j:
                    mid_row = max(i + 1, l[j])
                    top_end = mid_row - 1
                    bot_start = mid_row + 1

                    w2 = min(w1, w3) - 1
                    if w2 > 0:
                        area = (top_end - top_start + 1) * w1 + w2 + (bot_end - bot_start + 1) * w3
                        if area > best_area:
                            best_area = area
                            best = (col,
                                    top_start, top_end, w1,
                                    mid_row, mid_row, w2,
                                    bot_start, bot_end, w3)

                # --- Case 2: l[j] <= i (symmetric 1-row middle) ---
                if l[j] <= i:
                    mid_row = min(j - 1, rgt[i])
                    top_end = mid_row - 1
                    bot_start = mid_row + 1

                    w2 = min(w1, w3) - 1
                    if w2 > 0:
                        area = (top_end - top_start + 1) * w1 + w2 + (bot_end - bot_start + 1) * w3
                        if area > best_area:
                            best_area = area
                            best = (col,
                                    top_start, top_end, w1,
                                    mid_row, mid_row, w2,
                                    bot_start, bot_end, w3)

                # --- Case 3: r[i] <= l[j] (true middle band) ---
                if rgt[i] <= l[j]:
                    top_end = rgt[i] - 1
                    bot_start = l[j] + 1

                    mid_start = rgt[i]
                    mid_end = l[j]

                    min_middle = min_h[mid_start][mid_end]
                    w2 = min(w1 - 1, w3 - 1, min_middle)

                    if w2 > 0:
                        area = ((top_end - top_start + 1) * w1
                                + (mid_end - mid_start + 1) * w2
                                + (bot_end - bot_start + 1) * w3)
                        if area > best_area:
                            best_area = area
                            best = (col,
                                    top_start, top_end, w1,
                                    mid_start, mid_end, w2,
                                    bot_start, bot_end, w3)

    # No solution
    if best_area < 0:
        sys.stdout.write("-1\n")
        return

    # Paint the best solution onto the grid (mark with 8)
    (col,
     r1s, r1e, w1,
     r2s, r2e, w2,
     r3s, r3e, w3) = best

    for rr in range(r1s, r1e + 1):
        for cc in range(col, col + w1):
            grid[rr][cc] = 8

    for rr in range(r2s, r2e + 1):
        for cc in range(col, col + w2):
            grid[rr][cc] = 8

    for rr in range(r3s, r3e + 1):
        for cc in range(col, col + w3):
            grid[rr][cc] = 8

    # Output
    out_lines = [str(best_area)]
    for r in range(n):
        out_lines.append(" ".join(map(str, grid[r])))
    sys.stdout.write("\n".join(out_lines) + "\n")


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

---

## 5) Compressed editorial

- Precompute `dist[r][c]`: consecutive empty cells to the right from each cell (`0`→positive, `1`→0).
- Fix a left column `c`. Define histogram `height[r]=dist[r][c]` (max possible width starting at `c` in row `r`).
- Compute nearest smaller indices `l[r], r[r]` with monotonic stacks (classic histogram technique).
- Enumerate two critical rows `i` (top width `w1=height[i]`) and `j` (bottom width `w3=height[j]`, with `j≥i+2`).
- Middle width must be `w2 < min(w1,w3)` and fit its rows; handle 3 positional cases:
  1) `r[i] ≥ j`: place a 1-row middle; `w2=min(w1,w3)-1`.
  2) `l[j] ≤ i`: symmetric to (1).
  3) `r[i] ≤ l[j]`: middle spans rows `[r[i]..l[j]]`; `w2=min(w1-1,w3-1,min(height over middle))`.
- Precompute `min(height[L..R])` via `O(N^2)` DP table for O(1) middle-band min.
- Track best area and reconstruction info, paint chosen cells with `8`.
- Total complexity `O(M*N^2)`; `N,M≤180` fits.

