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
", 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 = 0, s = 1, c = gc;
    while (c <= 32)    c = gc;
    if (c == '-')    s = -1, c = gc;
    for (; isdigit(c); c = gc)
        x = x * 10 + c - 48;
    return x * s;
}

const int maxn = 1e5 + 5;

vector<ll> getLength(string s) {
    vector<ll> v(26, 0);
    int len = 1, last = -1;

    rep(i, 0, s.length() - 1) {
        int now = s[i] - 'a';
        if (now == last) {
            len++;
        } else {
            last = now;
            len = 1;
        }
        v[now] = max(v[now], (ll)len);
    }

    return v;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);

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

    rep(i, 1, n - 1) {
        cin >> s;
        int prec = s[0] - 'a', sufc = s.back() - 'a';
        int prel = 1, sufl = 1;
        rep(j, 1, s.length() - 1) {
            if (s[j] - 'a' == prec)    prel++;
            else    break;
        }
        irep(j, s.length() - 2, 0) {
            if (s[j] - 'a' == sufc)    sufl++;
            else    break;
        }

        vector<ll> cur = getLength(s);
        if (prel == s.length()) {
            rep(j, 0, 25) {
                if (j == prec) {
                    Len[j] += (Len[j] + 1) * prel;
                } else {
                    Len[j] = min(Len[j], 1ll);
                }
            }
        } else {
            rep(j, 0, 25) {
                if (Len[j])
                    Len[j] = (prec == j) * prel + (sufc == j) * sufl + 1;
                Len[j] = max(Len[j], cur[j]);
            }
        }

        rep(j, 0, 25)    Len[j] = min(Len[j], (ll)1e9);
    }

    ll ans = -1;
    rep(i, 0, 25)    ans = max(ans, Len[i]);
    cout << ans << endl;
    return 0;
}