其实这题可以直接使用 BFS。

AC code：

#include <bits/stdc++.h>
using namespace std;
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};

int main() {
    freopen("honey1.in", "r", stdin);
    freopen("honey1.out", "w", stdout);
    int T;
    cin >> T;
    while (T--) {
        int k;
        cin >> k;
        vector<string> grid(k);
        int start_x = -1, start_y = -1;
        for (int i = 0; i < k; ++i) {
            cin >> grid[i];
            for (int j = 0; j < k; ++j) {
                if (grid[i][j] == 'B') {
                    start_x = i;
                    start_y = j;
                }
            }
        }
        vector<vector<bool> > visited(k, vector<bool>(k, false));
        queue<pair<int, int> > q;
        q.push({start_x, start_y});
        visited[start_x][start_y] = true;
        bool found = false;
        while (!q.empty() && !found) {
            auto [x, y] = q.front();
            q.pop();
            for (int d = 0; d < 4; ++d) {
                int nx = x + dx[d];
                int ny = y + dy[d];
                if (nx < 0 || nx >= k || ny < 0 || ny >= k) {
                    continue;
                }
                if (visited[nx][ny]) {
                    continue;
                }
                if (grid[nx][ny] == 'L') {
                    found = true;
                    break;
                }
                if (grid[nx][ny] == 'R' || grid[nx][ny] == 'B') {
                    visited[nx][ny] = true;
                    q.push({nx, ny});
                }
            }
        }
        cout << (found ? "Yes" : "No") << endl;
    }
    return 0;
}