CF #541div2 E

题目本质：忽略串的变化，只记载26个字母的相关变化。

解决方法：

在上一次与本次的转移过程中，情况并不多，主要取决于本次串的首尾字母，若不是本次的首尾字母，会被置1；如果是的话，分情况接一下并更新。另外应该不会忘记的就是要拿本次串的最长串再更新一下。

复杂度最差时不是O（n²）吗？数据骗了。

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <ctime>
 #include <cctype>
 #include <climits>
 #include <iostream>
 #include <iomanip>
 #include <algorithm>
 #include <string>
 #include <sstream>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #include <vector>
 #include <list>
 #include <fstream>
 #define ri readint()
 #define gc getchar()
 #define R(x) scanf("%d", &x)
 #define W(x) printf("%d\n", x)
 #define init(a, b) memset(a, b, sizeof(a))
 #define rep(i, a, b) for (int i = a; i <= b; i++)
 #define irep(i, a, b) for (int i = a; i >= b; i--)
 #define ls p << 1
 #define rs p << 1 | 1
 using namespace std;

 typedef double db;
 typedef long long ll;
 typedef unsigned long long ull;
 typedef pair<int, int> P;
 const int inf = 0x3f3f3f3f;
 const ll INF = 1e18;

 inline int readint() {
     int x = , s = , c = gc;
     while (c <= )    c = gc;
     if (c == '-')    s = -, c = gc;
     for (; isdigit(c); c = gc)
         x = x *  + c - ;
     return x * s;
 }

 const int maxn = 1e5 + ;

 vector<ll> getLength(string s) {
     vector<ll> v(, );
     int len = , last = -;

     rep(i, , s.length() - ) {
         int now = s[i] - 'a';
         if (now == last) {
             len++;
         } else {
             last = now;
             len = ;
         }
         v[now] = max(v[now], (ll)len);
     }

     return v;
 }

 int main() {
     ios_base::sync_with_stdio(false);
     cin.tie();

     int n;
     cin >> n;
     string s;
     cin >> s;
     vector<ll> Len = getLength(s);

     rep(i, , n - ) {
         cin >> s;
         int prec = s[] - 'a', sufc = s.back() - 'a';
         int prel = , sufl = ;
         rep(j, , s.length() - ) {
             if (s[j] - 'a' == prec)    prel++;
             else    break;
         }
         irep(j, s.length() - , ) {
             if (s[j] - 'a' == sufc)    sufl++;
             else    break;
         }

         vector<ll> cur = getLength(s);
         if (prel == s.length()) {
             rep(j, , ) {
                 if (j == prec) {
                     Len[j] += (Len[j] + ) * prel;
                 } else {
                     Len[j] = min(Len[j], 1ll);
                 }
             }
         } else {
             rep(j, , ) {
                 if (Len[j])
                     Len[j] = (prec == j) * prel + (sufc == j) * sufl + ;
                 Len[j] = max(Len[j], cur[j]);
             }
         }

         rep(j, , )    Len[j] = min(Len[j], (ll)1e9);
     }

     ll ans = -;
     rep(i, , )    ans = max(ans, Len[i]);
     cout << ans << endl;
     return ;
 }