CF #541 E. String Multiplication

题意：

　　 给定一系列字符串，每次都是后一个字符串和前面的融合，这个融合操作就是原来的串分成独立的，然后把新串插入到这些空格中。问最后，最长的相同连续的长度。

思路：

　　这道题可以贪心的来，我们压缩状态，记录串中每个字母对应最长的长度。然后分类讨论处理就行了。

#include <algorithm>
#include <iterator>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <iomanip>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <stack>
#include <cmath>
#include <queue>
#include <list>
#include <map>
#include <set>
#include <cassert>

using namespace std;
#define lson (l, mid, rt << 1)
#define rson (mid + 1, r, rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost                    \
    ios::sync_with_stdio(false); \
    cin.tie(0)
#define rep(a, b, c) for (int a = (b); a <= (c); ++a)
#define max3(a, b, c) max(max(a, b), c);
#define min3(a, b, c) min(min(a, b), c);

const ll oo = 1ll << 17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000; //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9 + 7;
const double esp = 1e-8;
const double PI = acos(-1.0);
const double PHI = 0.61803399; //黄金分割点
const double tPHI = 0.38196601;

template <typename T>
inline T read(T &x) {
    x = 0;
    int f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    return x = f ? -x : x;
}

inline void cmax(int &x, int y) {
    if (x < y) x = y;
}
inline void cmax(ll &x, ll y) {
    if (x < y) x = y;
}
inline void cmin(int &x, int y) {
    if (x > y) x = y;
}
inline void cmin(ll &x, ll y) {
    if (x > y) x = y;
}

/*-----------------------showtime----------------------*/
const ll big = 1e9 + 3;
ll dp[30], tmp[30];
string str;
int main() {
    int n;
    scanf("%d", &n);
    ll ans = 0;
    rep(cc, 1, n) {
        cin >> str;

        int len = str.length();
        int flag = 1;
        for (int i = 0; i < len; i++)
            if (str[i] != str[0]) flag = 0;

        if (flag) {
            int id = (str[0] - 'a');
            rep(i, 0, 25) {
                if (i == id) continue;
                if (dp[i]) dp[i] = 1;
            }

            if (dp[id]) {
                ll t = min(big, 1ll * (dp[id] + 1) * len + dp[id]);
                dp[id] = max(dp[id], t);
            }
            dp[id] = max(1ll * len, dp[id]);
        } else {
            rep(i, 0, 25) {
                if (dp[i]) dp[i] = 1;
                tmp[i] = 0;
            }
            ll e = 1;
            char la = str[0];
            str += "A";
            rep(i, 1, len) {
                if (str[i] != la) {
                    int id = (int)(la - 'a');
                    tmp[id] = max(tmp[id], e);
                    //  debug(e);
                    la = str[i];
                    e = 1;
                } else
                    e++;
            }
            ll c1 = 0, c2 = 0;
            for (int i = 0; i < len && str[i] == str[0]; i++) c1++;
            for (int i = len - 1; i >= 0 && str[i] == str[len - 1]; i--) c2++;

            if (str[0] == str[len - 1]) {
                int id = (int)(str[0] - 'a');
                if (dp[id]) dp[id] = min(big, max(dp[id], 1ll + c1 + c2));
            } else {
                int id = (int)(str[0] - 'a');
                if (dp[id]) dp[id] = min(big, max(dp[id], 1ll + c1));
                id = (int)(str[len - 1] - 'a');
                if (dp[id]) dp[id] = min(big, max(dp[id], 1ll + c2));
            }
            rep(i, 0, 25) {
                dp[i] = max(dp[i], tmp[i]);
            }
        }
    }

    rep(i, 0, 25) ans = max(ans, dp[i]);
    printf("%lld\n", ans);
    return 0;
}