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>

/*

⊂_ヽ
　 ＼＼ Λ＿Λ  来了老弟
　　 ＼('ㅅ')
　　　 >　⌒ヽ
　　　/ 　 へ＼
　　 /　　/　＼＼
　　 ﾚ　ノ　　 ヽ_つ
　　/　/
　 /　/|
　(　(ヽ
　|　|、＼
　| 丿 ＼ ⌒)
　| |　　) /
'ノ )　　Lﾉ

*/

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#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 '
'

#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
", ans);
            return 0;
}