## 1) Abridged problem statement

You are given an `n × m` chessboard (`1 ≤ n,m ≤ 15`) containing:
- `*` : the only white piece, a **king** (moves 1 square in any of 8 directions),
- black pieces: `K` (knight), `B` (bishop), `R` (rook),
- `.` empty squares.

Black pieces **do not move**. The white king must capture **all** black pieces, one by one, moving like a king. The king is **never allowed to move to a square that is attacked by any black piece that is still on the board**. When the king captures a black piece, that piece is removed.

Find the **minimum number of king moves** to capture all black pieces, or `-1` if impossible. If there are no black pieces, output `0`.

Constraint: total number of pieces (including the king) is at most 15.

---

## 2) Detailed editorial (how the given solution works)

### Key observation: black pieces are static, only "alive/dead" matters
Since black pieces never move, the only changing parts of the game are:
1. the **current position** of the white king, and
2. which black pieces have already been **captured** (removed).

So we can model the problem as a shortest path in a state graph.

### State representation
Let there be `k` black pieces (`k ≤ 14` since total pieces ≤ 15 and one is the king).

A state is:
- `(r, c, mask)`
where:
- `(r, c)` is the king's current cell,
- `mask` is a `k`-bit bitmask, where bit `i = 1` means black piece `i` has been captured.

Goal state: any `(r, c, (1<<k)-1)` (all captured).

Start state: king's initial location with `mask = 0`.

### Precompute attacked squares per piece (ignoring captures for now)
The solution precomputes, for each board cell `(r,c)`, which pieces would attack it **if they are still alive**:
- `attack[r][c]` is a set of indices of black pieces that attack `(r,c)`.

How each black piece attacks:
- **Knight (`K`)**: the 8 L-shaped squares.
- **Bishop (`B`)**: rays along diagonals until board edge or until a non-`.` cell blocks further squares.
- **Rook (`R`)**: rays along ranks/files until board edge or blocked.

Important nuance:
- Blocking is computed using the **initial board contents** (`tbl`).
- This means rays stop at the first non-empty cell, which can be `*`, `K`, `B`, or `R`.

Why is this still workable here?
- The code later checks attacks only from **alive** pieces (`mask`).
- Captured pieces are treated as removed for "can attack?" purposes.
- (Note: A fully rigorous chess-physics model would also update sliding lines when a blocking piece is captured. This solution assumes the precomputed blocking is sufficient for the intended problem constraints/interpretation and matches the provided file's approach.)

### BFS over extended states
All moves cost 1, so we use BFS.

From state `(r,c,mask)`:
- Try moving the king to each of its 8 neighboring squares `(nr,nc)` (inside bounds).
- The move is legal only if `(nr,nc)` is **not attacked** by any **alive** piece:
  - look at all `idx` in `attack[nr][nc]`,
  - if any such `idx` has `(mask & (1<<idx)) == 0`, then that attacker is alive → square is forbidden.
- If the destination contains a black piece, we update the mask to include that piece (captured).

We store distances in:
- `dist[n][m][1<<k]`, initialized to `-1`.

The first time we pop a state with `mask == all_ones`, we output its distance (minimum moves).

If BFS ends without reaching the goal: output `-1`.

### Complexity
- Board size: up to `15×15 = 225`.
- `k ≤ 14` ⇒ masks up to `2^14 = 16384`.
- Total states: `225 * 16384 ≈ 3.7 million` (upper bound).
- Each state explores up to 8 moves; attack checking iterates over a set per cell (usually small).

This fits within limits in C++.

---

## 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;
vector<pair<int, int>> pieces;
vector<char> ptype;
int kpos;

void read() {
    cin >> n >> m;
    tbl.resize(n);
    cin >> tbl;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            if(tbl[i][j] == '*') {
                kpos = i * m + j;
            } else if(tbl[i][j] == 'K' || tbl[i][j] == 'B' || tbl[i][j] == 'R') {
                pieces.emplace_back(i, j);
                ptype.push_back(tbl[i][j]);
            }
        }
    }
}

void solve() {
    // We can solve this with a fairly simple BFS by extending the state. As
    // black pieces don't move, the state is fully defined by the position of
    // the king, and the state of figures that have been taken (which can be
    // kept as a 3-bit mask). Initially we will just cover the cells that are
    // blocked.

    int k = pieces.size();
    if(k == 0) {
        cout << 0 << "\n";
        return;
    }

    vector<vector<set<int>>> attack(n, vector<set<int>>(m));
    for(int idx = 0; idx < k; idx++) {
        auto [ci, cj] = pieces[idx];
        char type = ptype[idx];
        if(type == 'K') {
            for(int di = -2; di <= 2; di++) {
                for(int dj = -2; dj <= 2; dj++) {
                    if(abs(di) + abs(dj) == 3) {
                        int ni = ci + di, nj = cj + dj;
                        if(ni >= 0 && ni < n && nj >= 0 && nj < m) {
                            attack[ni][nj].insert(idx);
                        }
                    }
                }
            }
        } else if(type == 'B') {
            for(int d = 0; d < 4; d++) {
                int di = (d == 0 || d == 1) ? 1 : -1;
                int dj = (d == 0 || d == 2) ? 1 : -1;
                for(int step = 1;; step++) {
                    int ni = ci + di * step, nj = cj + dj * step;
                    if(ni < 0 || ni >= n || nj < 0 || nj >= m) {
                        break;
                    }
                    attack[ni][nj].insert(idx);
                    if(tbl[ni][nj] != '.') {
                        break;
                    }
                }
            }
        } else if(type == 'R') {
            for(int d = 0; d < 4; d++) {
                int di = (d == 0) ? 1 : ((d == 1) ? -1 : 0);
                int dj = (d == 2) ? 1 : ((d == 3) ? -1 : 0);
                for(int step = 1;; step++) {
                    int ni = ci + di * step, nj = cj + dj * step;
                    if(ni < 0 || ni >= n || nj < 0 || nj >= m) {
                        break;
                    }
                    attack[ni][nj].insert(idx);
                    if(tbl[ni][nj] != '.') {
                        break;
                    }
                }
            }
        }
    }

    queue<array<int, 3>> q;
    vector<vector<vector<int>>> dist(
        n, vector<vector<int>>(m, vector<int>(1 << k, -1))
    );
    int start_mask = 0;
    dist[kpos / m][kpos % m][start_mask] = 0;
    q.push({kpos / m, kpos % m, start_mask});

    int dr[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
    int dc[8] = {-1, 0, 1, -1, 1, -1, 0, 1};

    while(!q.empty()) {
        auto [ci, cj, mask] = q.front();
        q.pop();

        if(mask == (1 << k) - 1) {
            cout << dist[ci][cj][mask] << "\n";
            return;
        }

        for(int d = 0; d < 8; d++) {
            int ni = ci + dr[d];
            int nj = cj + dc[d];
            if(ni < 0 || ni >= n || nj < 0 || nj >= m) {
                continue;
            }

            bool attacked = false;
            for(int idx: attack[ni][nj]) {
                if(!(mask & (1 << idx))) {
                    attacked = true;
                    break;
                }
            }
            if(attacked) {
                continue;
            }

            int nmask = mask;
            char cell = tbl[ni][nj];
            if(cell == 'K' || cell == 'B' || cell == 'R') {
                for(int idx = 0; idx < k; idx++) {
                    if(pieces[idx] == make_pair(ni, nj)) {
                        nmask |= (1 << idx);
                        break;
                    }
                }
            }

            if(dist[ni][nj][nmask] == -1) {
                dist[ni][nj][nmask] = dist[ci][cj][mask] + 1;
                q.push({ni, nj, nmask});
            }
        }
    }

    cout << -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 (same approach) with detailed comments

```python
from collections import deque

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

    pieces = []   # list of (r,c) for each black piece
    ptype = []    # 'K','B','R' for each piece (black knight/bishop/rook)
    king_r = king_c = -1

    # Parse the board: find king and black pieces
    for i in range(n):
        for j in range(m):
            ch = tbl[i][j]
            if ch == '*':
                king_r, king_c = i, j
            elif ch in "KBR":
                pieces.append((i, j))
                ptype.append(ch)

    k = len(pieces)
    if k == 0:
        print(0)
        return

    # attack[r][c] will store a list of indices of pieces that attack (r,c).
    # Using list-of-lists (instead of set) is fine; duplicates won't happen here.
    attack = [[[] for _ in range(m)] for _ in range(n)]

    # Precompute all attacked squares for each black piece, based on initial board.
    for idx, ((ci, cj), typ) in enumerate(zip(pieces, ptype)):
        if typ == 'K':  # black knight
            for di in range(-2, 3):
                for dj in range(-2, 3):
                    if abs(di) + abs(dj) == 3:  # (2,1) / (1,2)
                        ni, nj = ci + di, cj + dj
                        if 0 <= ni < n and 0 <= nj < m:
                            attack[ni][nj].append(idx)

        elif typ == 'B':  # bishop: 4 diagonal rays
            dirs = [(1, 1), (1, -1), (-1, 1), (-1, -1)]
            for di, dj in dirs:
                step = 1
                while True:
                    ni, nj = ci + di * step, cj + dj * step
                    if not (0 <= ni < n and 0 <= nj < m):
                        break
                    attack[ni][nj].append(idx)
                    # Stop ray when hitting a non-empty square
                    if tbl[ni][nj] != '.':
                        break
                    step += 1

        else:  # 'R' rook: 4 orthogonal rays
            dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]
            for di, dj in dirs:
                step = 1
                while True:
                    ni, nj = ci + di * step, cj + dj * step
                    if not (0 <= ni < n and 0 <= nj < m):
                        break
                    attack[ni][nj].append(idx)
                    if tbl[ni][nj] != '.':
                        break
                    step += 1

    # Map piece position -> index to quickly set capture bit
    pos_to_idx = {pos: i for i, pos in enumerate(pieces)}

    # dist[r][c][mask] = min moves, initialized to -1 (unvisited)
    # Sizes: n * m * (1<<k)
    dist = [[[-1] * (1 << k) for _ in range(m)] for _ in range(n)]

    start_mask = 0
    dist[king_r][king_c][start_mask] = 0
    q = deque([(king_r, king_c, start_mask)])

    # 8 king moves
    dr = [-1, -1, -1,  0, 0, 1, 1, 1]
    dc = [-1,  0,  1, -1, 1,-1, 0, 1]

    full = (1 << k) - 1

    while q:
        r, c, mask = q.popleft()
        curd = dist[r][c][mask]

        if mask == full:
            print(curd)
            return

        for t in range(8):
            nr, nc = r + dr[t], c + dc[t]
            if not (0 <= nr < n and 0 <= nc < m):
                continue

            # Check if (nr,nc) is attacked by any alive piece
            attacked = False
            for idx in attack[nr][nc]:
                if (mask & (1 << idx)) == 0:  # piece still alive
                    attacked = True
                    break
            if attacked:
                continue

            nmask = mask
            # If we moved onto a black piece square, we capture it
            if (nr, nc) in pos_to_idx:
                nmask |= 1 << pos_to_idx[(nr, nc)]

            if dist[nr][nc][nmask] == -1:
                dist[nr][nc][nmask] = curd + 1
                q.append((nr, nc, nmask))

    print(-1)


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

---

## 5) Compressed editorial

Model as shortest path with BFS over states `(king_row, king_col, captured_mask)`, where `captured_mask` indicates which black pieces are removed. Precompute for each cell the set of black piece indices that attack it (knight moves, bishop/rook rays blocked by first non-empty cell). In BFS, a king move to a neighbor is allowed only if that neighbor is not attacked by any piece whose bit is still 0 in the mask. Moving onto a piece sets its bit (capture). Distance over `n*m*2^k` states gives the minimum moves; if unreachable output `-1`, if no pieces output `0`.
