题解 ABC334E【Christmas Color Grid 1】

先求出初始时绿连通块数量。

枚举每个红色格子，将其染成绿色本应增加一个绿连通块，但是它每与一个绿连通块相邻，就又会减少一个绿连通块。根据上述规则可以求出每个红色格子染绿后的绿连通块数量，求平均值即可。

时间复杂度 \(O(nm)\)。

// Problem: E - Christmas Color Grid 1
// Contest: AtCoder - UNIQUE VISION Programming Contest 2023 Christmas (AtCoder Beginner Contest 334)
// URL: https://atcoder.jp/contests/abc334/tasks/abc334_e
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(int x = (y); x <= (z); ++x)
#define per(x, y, z) for(int x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;

mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {
    uniform_int_distribution<int> dist(L, R);
    return dist(rnd);
}

template<typename T> void chkmin(T& x, T y) {if(x > y) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}

template<int mod>
inline unsigned int down(unsigned int x) {
	return x >= mod ? x - mod : x;
}

template<int mod>
struct Modint {
	unsigned int x;
	Modint() = default;
	Modint(unsigned int x) : x(x) {}
	friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}
	friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}
	friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}
	friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}
	friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}
	friend Modint operator/(Modint a, Modint b) {return a * ~b;}
	friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}
	friend Modint operator~(Modint a) {return a ^ (mod - 2);}
	friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}
	friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}
	friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}
	friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}
	friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}
	friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}
	friend Modint& operator++(Modint& a) {return a += 1;}
	friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}
	friend Modint& operator--(Modint& a) {return a -= 1;}
	friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}
	friend bool operator==(Modint a, Modint b) {return a.x == b.x;}
	friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};

typedef Modint<998244353> mint;

const int N = 1e3 + 5;
const int nxt[4][2] = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};

int n, m, k, vis[N][N];
string s[N];
mint cnt, ans;

void dfs(int x, int y, int u) {
    vis[x][y] = u;
    rep(d, 0, 3) {
        int nx = x + nxt[d][0], ny = y + nxt[d][1];
        if(s[nx][ny] == '#' && !vis[nx][ny]) dfs(nx, ny, u);
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    cin >> n >> m;
    s[0] = s[n + 1] = string(m + 2, ' ');
    rep(i, 1, n) {
        cin >> s[i];
        s[i] = ' ' + s[i] + ' ';
    }
    rep(i, 1, n) rep(j, 1, m) if(s[i][j] == '#' && !vis[i][j]) dfs(i, j, ++k);
    rep(i, 1, n) {
        rep(j, 1, m) {
            if(s[i][j] == '.') {
                ++cnt;
                set<int> st;
                rep(d, 0, 3) {
                    int nx = i + nxt[d][0], ny = j + nxt[d][1];
                    if(s[nx][ny] == '#') st.insert(vis[nx][ny]);
                }
                ans += k - (int)st.size() + 1;
            }
        }
    }
    cout << ans / cnt << endl;
    return 0;
}