题解（带批注）：

#include <iostream>
#include <vector>
#include <queue>
#include <set>
#include <iomanip>
#include <string>

using i64 = long long;

constexpr int dirs[][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

int main() {
stdiossync_with_stdio(false);
std::cin.tie(nullptr);

int n, m;
std::cin >> n >> m;

std::vector<std::string> grid(n);
for (auto &row : grid)
    std::cin >> row;

std::vector<std::vector<int>> color(n, std::vector<int>(m, 0));
int k = 0;

// BFS标记蓝色连通分量
for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
        if (grid[i][j] != 'B' || color[i][j]) continue;
        
        std::queue<std::pair<int, int>> q;
        q.emplace(i, j);
        color[i][j] = ++k;
        
        while (!q.empty()) {
            auto [r, c] = q.front(); 
            q.pop();
            
            for (auto &[dr, dc] : dirs) {
                int nr = r + dr, nc = c + dc;
                if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
                if (grid[nr][nc] != 'B' || color[nr][nc]) continue;
                
                color[nr][nc] = k;
                q.emplace(nr, nc);
            }
        }
    }
}

i64 numerator = 0, denominator = 0;

// 统计白色格子周围的蓝色分量
for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
        if (grid[i][j] != 'W') continue;
        
        std::set<int> adjacentColors;
        for (auto &[dr, dc] : dirs) {
            int nr = i + dr, nc = j + dc;
            if (nr < 0 || nr >= n || nc < 0 || nc >= m) continue;
            if (grid[nr][nc] == 'B') {
                adjacentColors.insert(color[nr][nc]);
            }
        }
        
        numerator += k - adjacentColors.size() + 1;
        denominator += 1;
    }
}

// 精确输出，避免格式错误
if (denominator == 0) {
    std::cout << "0.000000000000\n";
} else {
    double result = static_cast<double>(numerator) / denominator;
    std::cout << std::fixed << std::setprecision(12) << result << "\n";
}

return 0;

}