除#1 #2 外全部WA，20 pts状压求调

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;
typedef long long LL;
const int N = 103, M = 10, S = 1 << M;
int n, m, g[N], cnt[S];
vector<int> states;
vector<int> head[S];
LL f[2][S][S];
int main() {
	cin >> n >> m;
	for(int i = 1; i <= n; i ++)
		for(int j = 0; j < m; j ++) {
			char c; cin >> c;
			if(c == 'H') g[i] += 1 << j;
		}
	for(int s = 0; s < (1 << m); s ++)
		if(!(s & s >> 1 || s & s >> 2)) {
			states.push_back(s);
			int s_ = s, tot = 0;
			while(s_) tot += s_ & 1, s_ >>= 1;
			cnt[s] = tot;
		}
	for(int i = 0; i < (int)states.size(); i ++)
		for(int j = 0; j < (int)states.size(); j ++) {
			int s1 = states[i], s2 = states[j];
			if(!(s1 & s2)) head[s1].push_back(s2);
		}
	for(int i = 1; i <= n + 2; i ++)
		for(int a = 0; a < (int)states.size(); a ++) {
			int s1 = states[a];
			if(!(g[i] & s1)) {
				for(int b = 0; b < (int)head[s1].size(); b ++) {
					int s2 = states[b];
					for(int c = 0; c < (int)head[s2].size(); c ++) {
						int s3 = states[c];
						if(!(s1 & s3))
							f[i & 1][s1][s2] = max(f[i & 1][s1][s2], f[(i - 1) & 1][s2][s3] + cnt[s1]);
					}
				}
			}
		}
	cout << f[(n + 2) & 1][0][0];
	return 0;
}