传送门

又是神仙DP

发现如果只有两个串就很好做了
于是这个神仙DP定义就从这里下手：令 $dp[p][c][l][r] 表示在 \([s_l, s_r]\) 这段字符串中，考虑从第 \(p\) 个位置开始的后缀，并要求这个字符至少为 \(c\)
考虑转移，因为这里有个「至少」，第一个转移是直接从 \(dp[p][c+1][l][r]\) 继承过来
考虑第二个转移
发现在 \([l, r]\) 中一定存在一个 \(c\) 到 \(c+1\) 的分界点，我们枚举这个分界点
同时如果我们强令为 \(c\) 的那些数与输入冲突了就break掉
于是这部分的转移方程为

\[dp[p][c][l][r] += \sum\limits_{i=l}^{i\leqslant r} dp[p+1][0][l][i]*dp[p][c+1][i+1][r] \]

这样就（在一定程度上）转化成了对两个字符串的处理
记得要把长度不够的串用‘a'-1补成一样长

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long 
#define ull unsigned long long
#define reg register int
//#define int long long 

int n;
char s[55][25];
int len[55], maxl;
const ll mod=990804011ll, base=131;
inline void md(ll& a, ll b) {a+=b; a=a>=mod?a-mod:a;}

namespace force{
	ull p[25];
	inline ull hashing(ull* h, int l, int r) {return h[r]-h[l-1]*p[r-l+1];}
	ll dfs(int u, char* t, ull ht, ull hs, bool lim) {
		//cout<<"dfs "<<u<<' '<<ht<<' '<<hs<<' '<<lim<<endl;
		if (u>n) return 1;
		ull h[25]; h[0]=0; bool full=1;
		//cout<<"len: "<<len[u]<<endl;
		for (reg i=1; i<=len[u]; ++i) {
			//cout<<"i: "<<i<<endl;
			if (s[u][i]=='?') {
				full=0; ll ans=0;
				//cout<<"pos1"<<endl;
				for (reg j=(lim?t[i]-'a'+1:0); j<26; ++j) {
					s[u][i]='a'+j; h[i]=h[i-1]*base+s[u][i];
					ans+=dfs(u, t, ht, h[i], lim&&(s[u][i]==t[i]));
				}
				s[u][i]='?';
				return ans;
			}
			else {
				h[i]=h[i-1]*base+s[u][i];
				if (lim && s[u][i]<t[i]) return 0;
				if (lim && s[u][i]!=t[i]) lim=0;
			}
		}
		if (full && !lim) return dfs(u+1, s[u], h[len[u]], 0, 1);
		return 0;
	}
	void solve() {
		printf("%lld\n", dfs(1, s[0], 0, 0, 0));
		exit(0);
	}
}

namespace task1{
	ull p[25];
	struct st{int u; ull ht, hs; bool lim; st(int a, ull b, ull c, bool d):u(a),ht(b),hs(c),lim(d){}};
	struct st_hash{inline size_t operator () (st a) const {return hash<ull>()(a.ht*a.hs);}};
	inline bool operator == (st a, st b) {return a.u==b.u&&a.ht==b.ht&&a.hs==b.hs&&a.lim==b.lim;}
	unordered_map<st, ll, st_hash> mp{5000, st_hash()};
	inline ull hashing(ull* h, int l, int r) {return h[r]-h[l-1]*p[r-l+1];}
	ll dfs(int u, char* t, ull ht, ull hs, bool lim) {
		//cout<<"dfs "<<u<<' '<<ht<<' '<<hs<<' '<<lim<<endl;
		if (u>n) return 1;
		st sit(u, ht, hs, lim);
		if (mp.find(sit)!=mp.end()) return mp[sit];
		ull h[25]; h[0]=0; bool full=1;
		//cout<<"len: "<<len[u]<<endl;
		for (reg i=1; i<=len[u]; ++i) {
			//cout<<"i: "<<i<<endl;
			if (s[u][i]=='?') {
				full=0; ll ans=0;
				//cout<<"pos1"<<endl;
				for (reg j=(lim?t[i]-'a'+1:0); j<26; ++j) {
					s[u][i]='a'+j; h[i]=h[i-1]*base+s[u][i];
					ans+=dfs(u, t, ht, h[i], lim&&(s[u][i]==t[i])), ans%=mod;
				}
				s[u][i]='?';
				mp[sit]=ans;
				return ans;
			}
			else {
				h[i]=h[i-1]*base+s[u][i];
				if (lim && s[u][i]<t[i]) {mp[sit]=0; return 0;}
				if (lim && s[u][i]!=t[i]) lim=0;
			}
		}
		if (full && !lim) {
			mp[sit]=dfs(u+1, s[u], h[len[u]], 0, 1);
			return mp[sit];
		}
		return 0;
	}
	void solve() {
		printf("%lld\n", dfs(1, s[0], 0, 0, 0)%mod);
		exit(0);
	}
}

namespace task{
	ll dp[25][30][52][52];
	ll dfs(int p, int c, int l, int r) {
		//cout<<"dfs "<<p<<' '<<c<<' '<<l<<' '<<r<<endl;
		if (~dp[p][c][l][r]) return dp[p][c][l][r];
		if (l>r) return dp[p][c][l][r]=1;
		if (p>maxl) return dp[p][c][l][r]=(l==r);
		if (c>26) return dp[p][c][l][r]=0;
		dp[p][c][l][r]=dfs(p, c+1, l, r);
		for (int i=l; i<=r; ++i) {
			//if (s[i][p]!=27&&s[i][p]!=c) break;
			if (!(s[i][p]==c || (s[i][p]==27&&c))) break;
			md(dp[p][c][l][r], dfs(p+1, 0, l, i)*dfs(p, c+1, i+1, r)%mod);
		}
		return dp[p][c][l][r];
	}
	void solve() {
		memset(dp, -1, sizeof(dp));
		printf("%lld\n", dfs(1, 0, 1, n));
		exit(0);
	}
}

signed main()
{
	scanf("%d", &n);
	for (reg i=1; i<=n; ++i) {
		scanf("%s", s[i]+1);
		len[i]=strlen(s[i]+1);
		maxl=max(maxl, len[i]);
		for (int j=1; j<=len[i]; ++j) 
			if (s[i][j]=='?') s[i][j]=27;
			else s[i][j]-='`';
	}
	//if (n*maxl<=10) force::solve();
	//else task1::solve();
	task::solve();
	
	return 0;
}