题意：找一个串中的最长连续两个回文子串长度

题解:建两个回文树，一个正着，一个反着，每次add之后last的长度就是后缀最长的回文串长度，然后两边加一遍即可

/**************************************************************    Problem: 2565    User: walfy    Language: C++    Result: Accepted    Time:164 ms    Memory:49928 kb****************************************************************/ //#pragma GCC optimize(2)//#pragma GCC optimize(3)//#pragma GCC optimize(4)//#pragma GCC optimize("unroll-loops")//#pragma comment(linker, "/stack:200000000")//#pragma GCC optimize("Ofast,no-stack-protector")//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")#include
    
     #define fi first#define se second#define db double#define mp make_pair#define pb push_back#define pi acos(-1.0)#define ll long long#define vi vector
     
      #define mod 998244353#define ld long double#define C 0.5772156649#define ls l,m,rt<<1#define rs m+1,r,rt<<1|1#define pll pair
      
       #define pil pair
       
        #define pli pair
        
         #define pii pair
         
          //#define cd complex
          
           #define ull unsigned long long#define base 1000000000000000000#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))#define fin freopen("a.txt","r",stdin)#define fout freopen("a.txt","w",stdout)#define fio ios::sync_with_stdio(false);cin.tie(0)template
           
            inline T const& MAX(T const &a,T const &b){return a>b?a:b;}template
            
             inline T const& MIN(T const &a,T const &b){return a
             
              =mod)a-=mod;}inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;} using namespace std; const double eps=1e-8;const ll INF=0x3f3f3f3f3f3f3f3f;const int N=200000+10,maxn=400000+10,inf=0x3f3f3f3f; struct PAM{ int ch[N][26],fail[N],cnt[N],len[N],s[N],ans[N]; int last,n,p; int newnode(int w) { for(int i=0;i<26;i++)ch[p][i] = 0; cnt[p] = 0; len[p] = w; return p++; } void init() { p = last = n = 0; newnode(0); newnode(-1); s[n] = -1; fail[0] = 1; } int getfail(int x) { while(s[n-len[x]-1] != s[n]) x = fail[x]; return x; } void add(int c) { s[++n] = c; int cur = getfail(last); if(!ch[cur][c]){ int now = newnode(len[cur]+2); fail[now] = ch[getfail(fail[cur])][c]; ch[cur][c] = now; } last = ch[cur][c]; ans[n]=len[last]; cnt[last]++; }}pam1,pam2;char s[N];int main(){ pam1.init(),pam2.init(); scanf("%s",s+1); int len=strlen(s+1); for(int i=1;i<=len;i++)pam1.add(s[i]-'a'); reverse(s+1,s+len+1); for(int i=1;i<=len;i++)pam2.add(s[i]-'a'); int res=0; for(int i=1;i