E - Colorings and Dominoes

DP + 算貢獻

首先觀察到每一行的貢獻和每一列的貢獻是獨立的，所以可以單獨算出每一行和每一列的貢獻之和

對某一行考慮

設 \(f[i][0]\) 為第 \(i\) 個元素可作為放骨牌的第一個位置的概率，\(f[i][1]\) 為第 \(i\) 個元素可作為放骨牌的第二個位置的概率

轉移：

\(f[i][0]=\frac12*(1-f[i-1][0])\)

\(f[i][1] = \frac12*f[i-1][0]\)

\(*\frac12\) 為只有一半的概率是該顏色

答案為 \(f[i][1]\) 之和 * 總方案數（\(2^{白色的方格數}\)

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;
typedef long long ll;
const int mod = 998244353;
const int N = 3e5 + 10;
ll f[N][2];
int n, m, cnt;
ll qmi(ll a, ll b)
{
	ll ans = 1;
	while(b)
	{
		if (b & 1)
			ans = ans * a % mod;
		 a = a * a % mod;
		 b >>= 1;
	}
	return ans % mod;
}

int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cin >> n >> m;
	vector<vector<char> > s(n + 10, vector<char> (m + 10));
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
		{
			cin >> s[i][j];
			if (s[i][j] == 'o')
				cnt++;
		}
	}
	ll ans = 0;
	ll tot = qmi(2, cnt), inv2 = qmi(2, mod - 2);
	for (int i = 1; i <= n; i++)
	{
		f[0][0] = 0;
		ll now = 0;
		for (int j = 1; j <= m; j++)
		{
			if (s[i][j] == '*')
			{
				f[j][0] = f[j][1] = 0;
				continue;
			}
			f[j][0] = inv2 * (1 - f[j-1][0] + mod) % mod;
			f[j][1] = inv2 * f[j-1][0] % mod;
			now = (now + f[j][1]) % mod;
		}
		ans = (ans + tot * now % mod) % mod;
	}
	
	for (int j = 1; j <= m; j++)
	{
		f[0][0] = 0;
		ll now = 0;
		for (int i = 1; i <= n; i++)
		{
			if (s[i][j] == '*')
			{
				f[i][0] = f[i][1] = 0;
				continue;
			}
			f[i][0] = inv2 * (1 - f[i-1][0] + mod) % mod;
			f[i][1] = inv2 * f[i-1][0] % mod;
			now = (now + f[i][1]) % mod;
		}
		ans = (ans + tot * now % mod) % mod;
	}
	cout << ans << endl;
	
	return 0;
}