## 1. Abridged problem statement

You are given an \(H \times W\) grid containing exactly 9 cells marked `X`, forming one 4-neighbor connected shape. You want to cut this shape into the minimum number \(K\) of **connected parts** (4-neighbor connectivity), then **shift** each part (translation only; no rotation/flip) so that the union becomes a **full 3×3 square**.

Output:
1) The minimum \(K\).
2) The original grid, but each `X` replaced by a letter `A..` indicating which part it belongs to.
3) A blank line.
4) A 3×3 grid showing the parts shifted to form the 3×3 square, using the same letters.

Any optimal solution is acceptable.

---

## 2. Detailed editorial

### Key observation: only 9 squares
There are exactly 9 occupied cells. The target 3×3 square also has exactly 9 cells. So the final arrangement is a **bijection** between the 9 input `X` cells and the 9 positions of the 3×3 square.

That suggests brute force over mappings:
- Number the 9 input cells \(0..8\).
- Number the 9 target positions \(0..8\) (where index \(p\) corresponds to target row \(p/3\), col \(p\%3\)).
- Consider a permutation `perm` of `0..8` where `perm[i]` is the target position assigned to input cell `i`.

There are \(9! = 362{,}880\) permutations, which is feasible.

### What makes cells belong to the same "part"?
A "part" is a connected set of input cells that will be shifted together by one translation vector \((\Delta r, \Delta c)\).

For a fixed permutation:
- Input cell \(i\) at \((r_i, c_i)\) is mapped to target \((R_i, C_i)\).
- The translation required for cell \(i\) is:
  \[
  \Delta r_i = R_i - r_i,\quad \Delta c_i = C_i - c_i
  \]
If two cells are in the **same part**, they must share the **same** translation vector (because the whole part is shifted rigidly with no rotation).

So a necessary condition for cells \(i\) and \(j\) to be in the same part is:
\[
(\Delta r_i, \Delta c_i) = (\Delta r_j, \Delta c_j)
\]

Additionally, each part must be connected **in the original logo**, using side-adjacency. So within the set of cells that share the same \((\Delta r, \Delta c)\), the number of connected components (considering adjacency in the original grid) equals the number of parts of that translation.

### Computing number of parts for a permutation
1. Precompute adjacency between the 9 cells in the **original** shape:
   - `adj[i]` contains `j` if cells \(i\) and \(j\) are 4-neighbor adjacent.

2. For a permutation:
   - Compute all \(\Delta r_i, \Delta c_i\).
   - Build a DSU (disjoint set union) on 9 nodes.
   - For each adjacent pair \((i, j)\) in the original shape:
     - If they also have the same \((\Delta r, \Delta c)\), then they can be in the same part → union them.

3. After processing all edges, each DSU component corresponds to **one connected part**.
   - Count DSU roots → number of parts for this permutation.

Take the minimum over all permutations.

### Constructing the output labeling
Once the best permutation is found:
- Rebuild the DSU.
- Assign letters `A, B, C, ...` to DSU components.
- Print:
  1) The original grid, replacing each `X` cell by its component letter.
  2) The 3×3 target grid, placing each original cell's letter at its assigned target position.

### Complexity
- \(9!\) permutations.
- For each permutation, work is \(O(9 + E)\) where \(E \le 12\) (edges among 9 grid cells are small).
- Total operations: a few million → fine.

---

## 3. C++ Solution

```cpp
#include <bits/stdc++.h>
// #include <coding_library/graph/dsu.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;
};

class DSU {
  public:
    int n;
    vector<int> par;
    vector<int> sz;

    DSU(int _n = 0) { init(_n); }

    void init(int _n) {
        n = _n;
        par.assign(n + 1, 0);
        sz.assign(n + 1, 0);
        for(int i = 0; i <= n; i++) {
            par[i] = i;
            sz[i] = 1;
        }
    }

    int root(int u) { return par[u] = ((u == par[u]) ? u : root(par[u])); }
    bool connected(int x, int y) { return root(x) == root(y); }

    int unite(int x, int y) {
        x = root(x), y = root(y);
        if(x == y) {
            return x;
        }
        if(sz[x] > sz[y]) {
            swap(x, y);
        }
        par[x] = y;
        sz[y] += sz[x];
        return y;
    }

    vector<vector<int>> components() {
        vector<vector<int>> comp(n + 1);
        for(int i = 0; i <= n; i++) {
            comp[root(i)].push_back(i);
        }
        return comp;
    }
};

int n, m;
vector<string> tbl;

void read() {
    cin >> n >> m;
    tbl.resize(n);
    cin >> tbl;
}

void solve() {
    // The main observation to make is that there are exactly 9 cells that are
    // an X, and 9! = 362880 is actually fairly small. This means we can try all
    // possible mappings to the 3x3 grid. Afterwards, we simply want to figure
    // out what's the least number of letters, such that the compositions would
    // be connected in both the original table and have the same "shapes". For a
    // given mapping, each cell has a delta (the shift from its original
    // position to its 3x3 position). Cells in the same part must share the same
    // delta (since parts are only shifted, not rotated). So we group cells by
    // delta, and within each group it's enough to count the connected
    // components (say using a DSU). The total number of components across
    // all groups is the number of parts for that permutation. We take the
    // minimum over all 9! permutations, where the total number of operations
    // will just be a few million and would fit in the constraints very
    // comfortably.

    vector<pair<int, int>> cells;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            if(tbl[i][j] == 'X') {
                cells.push_back({i, j});
            }
        }
    }

    vector<vector<int>> adj(9);
    for(int i = 0; i < 9; i++) {
        for(int j = i + 1; j < 9; j++) {
            if(abs(cells[i].first - cells[j].first) +
                   abs(cells[i].second - cells[j].second) ==
               1) {
                adj[i].push_back(j);
                adj[j].push_back(i);
            }
        }
    }

    auto build_dsu = [&](const vector<int>& p) {
        DSU dsu(9);
        vector<int> dr(9), dc(9);
        for(int i = 0; i < 9; i++) {
            dr[i] = p[i] / 3 - cells[i].first;
            dc[i] = p[i] % 3 - cells[i].second;
        }
        for(int i = 0; i < 9; i++) {
            for(int j: adj[i]) {
                if(j > i && dr[i] == dr[j] && dc[i] == dc[j]) {
                    dsu.unite(i, j);
                }
            }
        }
        return dsu;
    };

    vector<int> perm(9);
    iota(perm.begin(), perm.end(), 0);

    int best_parts = 10;
    vector<int> best_perm;

    do {
        auto dsu = build_dsu(perm);

        int cnt = 0;
        for(int i = 0; i < 9; i++) {
            if(dsu.root(i) == i) {
                cnt++;
            }
        }

        if(cnt < best_parts) {
            best_parts = cnt;
            best_perm = perm;
            if(best_parts == 1) {
                break;
            }
        }
    } while(next_permutation(perm.begin(), perm.end()));

    auto dsu = build_dsu(best_perm);

    vector<int> label(9, -1);
    int cur = 0;
    for(int i = 0; i < 9; i++) {
        int r = dsu.root(i);
        if(label[r] == -1) {
            label[r] = cur++;
        }
    }

    cout << best_parts << "\n";
    auto grid = tbl;
    for(int i = 0; i < 9; i++) {
        grid[cells[i].first][cells[i].second] = 'A' + label[dsu.root(i)];
    }
    for(auto& row: grid) {
        cout << row << "\n";
    }
    cout << "\n";

    vector<string> small(3, string(3, '.'));
    for(int i = 0; i < 9; i++) {
        small[best_perm[i] / 3][best_perm[i] % 3] = 'A' + label[dsu.root(i)];
    }
    for(auto& row: small) {
        cout << row << "\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();
        solve();
    }

    return 0;
}
```

---

## 4. Python Solution

```python
import sys
from itertools import permutations

# ---------- DSU (Union-Find) ----------
class DSU:
    def __init__(self, n: int):
        # We will use indices 0..n-1
        self.par = list(range(n))
        self.sz = [1] * n

    def root(self, x: int) -> int:
        # Path compression
        while self.par[x] != x:
            self.par[x] = self.par[self.par[x]]
            x = self.par[x]
        return x

    def unite(self, a: int, b: int) -> None:
        a, b = self.root(a), self.root(b)
        if a == b:
            return
        # Union by size: attach smaller to larger
        if self.sz[a] > self.sz[b]:
            a, b = b, a
        self.par[a] = b
        self.sz[b] += self.sz[a]


def solve() -> None:
    data = sys.stdin.read().strip().splitlines()
    h, w = map(int, data[0].split())
    grid = data[1:1+h]

    # Collect the 9 occupied cells
    cells = []
    for r in range(h):
        for c in range(w):
            if grid[r][c] == 'X':
                cells.append((r, c))

    # Build adjacency list among the 9 cells in the original figure
    adj = [[] for _ in range(9)]
    for i in range(9):
        r1, c1 = cells[i]
        for j in range(i + 1, 9):
            r2, c2 = cells[j]
            if abs(r1 - r2) + abs(c1 - c2) == 1:
                adj[i].append(j)
                adj[j].append(i)

    # Build DSU for a given mapping p[i] = target position (0..8) for input cell i
    def build_dsu(p):
        dsu = DSU(9)

        # Translation vectors needed for each cell i
        dr = [0] * 9
        dc = [0] * 9
        for i in range(9):
            tr, tc = divmod(p[i], 3)   # target row/col in 3x3
            r, c = cells[i]            # original position
            dr[i] = tr - r
            dc[i] = tc - c

        # Union adjacent original cells if they share the same translation
        for i in range(9):
            for j in adj[i]:
                if j > i and dr[i] == dr[j] and dc[i] == dc[j]:
                    dsu.unite(i, j)
        return dsu

    best_parts = 10
    best_perm = None

    # Try all 9! bijections from input cells to 3x3 target positions
    for p in permutations(range(9)):
        dsu = build_dsu(p)

        # Count components (roots)
        roots = set(dsu.root(i) for i in range(9))
        parts = len(roots)

        if parts < best_parts:
            best_parts = parts
            best_perm = p
            if best_parts == 1:
                break

    # Rebuild DSU for best permutation and assign letters to components
    dsu = build_dsu(best_perm)

    # Map DSU root -> label index 0..K-1
    root_to_label = {}
    next_label = 0
    for i in range(9):
        r = dsu.root(i)
        if r not in root_to_label:
            root_to_label[r] = next_label
            next_label += 1

    # Output K
    out = []
    out.append(str(best_parts))

    # Print original grid with X replaced by letters
    grid2 = [list(row) for row in grid]
    for i in range(9):
        r, c = cells[i]
        lab = root_to_label[dsu.root(i)]
        grid2[r][c] = chr(ord('A') + lab)
    out.extend("".join(row) for row in grid2)

    # Blank line
    out.append("")

    # Print 3x3 target arrangement
    small = [["."] * 3 for _ in range(3)]
    for i in range(9):
        tr, tc = divmod(best_perm[i], 3)
        lab = root_to_label[dsu.root(i)]
        small[tr][tc] = chr(ord('A') + lab)
    out.extend("".join(row) for row in small)

    sys.stdout.write("\n".join(out))


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

---

## 5. Compressed editorial

- There are exactly 9 `X` cells, and the goal is a filled 3×3 (also 9 cells).
- Try all \(9!\) bijections from the 9 input cells to the 9 target positions.
- For a mapping, each cell gets a translation \((\Delta r,\Delta c)\) from original to target. Cells can be in the same part iff they share the same translation, and the part must be connected in the original grid.
- Build adjacency among the 9 input cells (4-neighbor). For each mapping, DSU-union adjacent cells that also share the same translation. The number of DSU components is the number of parts for that mapping.
- Take the mapping with minimum components; label components with `A..`, print labeled original grid and the labeled 3×3 result.
