1. Abridged Problem Statement
Given an n×m grid (n,m≤25) with some cells marked '*' (rocks) and '.' (empty), you may drop a bomb at any cell (i,j). A bomb clears all rocks in row i and in column j. Compute the minimum number of bombs needed to destroy every rock.

2. Detailed Editorial

We need to choose some bombs (each picks one row and one column) so that every rock at (i,j) lies either in a chosen row or in a chosen column. If we drop k bombs, they involve at most k distinct rows and at most k distinct columns (you can reuse a row or column across bombs). Conversely, any choice of R rows and C columns can be realized with max(|R|,|C|) bombs by pairing up rows and columns (if one side is smaller, reuse arbitrary entries on the larger side).

Thus the problem reduces to this covering problem:
  - Select a subset R⊆{0…n−1} of rows and C⊆{0…m−1} of columns
  - For every rock at (i,j), we require i∈R or j∈C
  - The cost is max(|R|,|C|), and we want to minimize that.

We optimize by enumerating subsets of rows R. For a fixed R, any rock in a row of R is already covered; the remaining rocks lie in rows outside R. To cover them, we must include in C every column that contains at least one uncovered rock. Thus
  C = { j | ∃ i∉R with grid[i][j]=='*' }
and cost(R)=max(|R|, |C|). We try all 2^n subsets R, compute C by OR'ing bitmasks of the rows in the complement of R, and track the minimum cost.

To compute the union column-mask for any subset of rows efficiently, we build the array mask[s] = OR of single-row masks over all rows in set s, using the standard "lowbit" SOS-style merging:
  mask[i] = mask[i & -i] | mask[i ^ (i & -i)]
This fills all 2^n entries in O(2^n) time.

Time & memory:
  - There are 2^n subsets, n≤25 ⇒ ∼33 million.
  - For each subset, we do a few integer bit-operations (very fast).
  - Memory to store an array of size 2^n of ints is about 4×33M≈132 MB, within the 512 MB limit.

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<string> tbl;

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

void solve() {
    // Each bomb clears one full row and one full column at once, so if a plan
    // bombs r distinct rows and c distinct columns we can pair them up and only
    // need max(r, c) bombs. We enumerate every subset i of the n rows that we
    // choose to bomb: those rows are cleared completely, and the rocks in the
    // remaining rows must be killed by bombing their columns. mask[s] holds the
    // union (bitmask over columns) of rock columns across the rows in set s,
    // built by SOS-style merging of single-row masks. For each chosen row set i
    // the needed columns are mask of the complement, and the cost is
    // max(popcount(rows), popcount(columns)); we minimise this over all subsets.

    vector<int> mask(1 << n, 0);
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            if(tbl[i][j] == '*') {
                mask[1 << i] |= 1 << j;
            }
        }
    }

    for(int i = 1; i < (1 << n); i++) {
        int l = i & -i;
        mask[i] = mask[l] | mask[i ^ l];
    }

    int ans = n + m;
    for(int i = 0; i < (1 << n); i++) {
        int mask_m = mask[((1 << n) - 1) ^ i];
        ans = min(ans, max(__builtin_popcount(i), __builtin_popcount(mask_m)));
    }

    cout << ans << '\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

```python
import sys
input = sys.stdin.readline

def main():
    n, m = map(int, input().split())
    tbl = [input().strip() for _ in range(n)]

    # If there are no rocks, answer is 0.
    any_rock = any('*' in row for row in tbl)
    if not any_rock:
        print(0)
        return

    # To speed up, we enumerate on the smaller dimension (rows vs. columns).
    # If columns are fewer, we transpose the grid.
    transpose = False
    if m < n:
        transpose = True
        # Transpose tbl into size m x n
        new_tbl = [''.join(tbl[i][j] for i in range(n)) for j in range(m)]
        tbl = new_tbl
        n, m = m, n

    # Build single-row rock masks
    row_masks = []
    for i in range(n):
        mask = 0
        for j, ch in enumerate(tbl[i]):
            if ch == '*':
                mask |= 1 << j
        row_masks.append(mask)

    FULL = (1 << n) - 1
    # Precompute mask_union[s] = OR of row_masks over rows in subset s
    mask_union = [0] * (1 << n)
    # Initialize single-bit subsets
    for i in range(n):
        mask_union[1 << i] = row_masks[i]
    # Build up all subsets by lowbit trick
    for s in range(1, 1 << n):
        lowbit = s & -s
        if s != lowbit:
            mask_union[s] = mask_union[lowbit] | mask_union[s ^ lowbit]

    best = n + m
    # Enumerate all subsets of rows R (bitmask s)
    for s in range(1 << n):
        rows = s.bit_count()
        # Rocks not covered by these rows lie in rows in complement: FULL^s
        needed_cols = mask_union[FULL ^ s].bit_count()
        # cost = max(rows, needed_cols)
        cost = rows if rows >= needed_cols else needed_cols
        if cost < best:
            best = cost

    print(best)

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

Explanation of key points in the Python code:
- We optionally transpose if m<n so we always choose the smaller dimension for the 2^d enumeration.
- We represent each row by an integer bitmask of length m.
- We build a DP array `mask_union` of size 2^n, where `mask_union[s]` is the OR of all row-masks for rows in subset s. We do this with the "lowbit" trick in O(2^n) time.
- For each subset s of rows, uncovered rocks lie in rows not in s ⇒ columns needed = bitcount of `mask_union[complement]`. The answer is min over s of max(|s|, columns_needed).

5. Compressed Editorial

We want to cover all stars by choosing some rows R and columns C, minimizing max(|R|,|C|). For any R, let C be the set of columns that still contain rocks in rows ∉R. Then cost=max(|R|,|C|). Enumerate all 2^n subsets of rows, precompute per-subset column-masks by SOS-style DP on bitmasks, and take the minimum cost. This runs in O(2^n·n) time and O(2^n) space, which is fine for n≤25.
