Palindrome tree template title

Topic Link  https://www.luogu.org/problem/P3649

Title Description

Give you a string of lowercase Latin letters  S S. We define  s a s sub-string of the present value of the substring  s number of occurrences s multiplied by the length of the substring.

For a string you  S S, find the maximum value of all the presence of a palindromic substring.

#include <bits / STDC ++ H.>
 #define Met (A, B) Memset (A, B, the sizeof (A))
 #define LL Long Long
 #define ULL unsigned Long Long
 the using  namespace STD; 

const  int MAXN = 300005 ;
 const  int N = 26 is ;
 struct the PAM {
     int NE [MAXN] [N]; // Next pointer, next trie pointer and the like, pointing to the current string is the same at both ends of a character string constituting plus 
    int Fail [MAXN]; / / fail pointer to jump to the fail mismatch node pointer 
    int CNT [MAXN]; // represents the number obtained when (i contribution essentially different nodes represented by strings is not complete, the final count () function after running it again is correct) 
    int NUM [MAXN]; //Palindromic sequence represents the number of the rightmost point of the node i of the longest palindromic sequence represented by the end of the palindromic sequence 
    int len [MAXN]; // len [i] indicates the length of a palindromic sequence represented by node i (a node represents a palindromic sequence) 
    int S [MAXN]; // store the added character 
    int Last; // points to the node longest palindromic sequence represented by a letter to add the new formed. 
    int n-; // represents the number of characters to add. 
    int P; // represents the number of nodes added. 
    int newNode ( int L) { // create a new node 
        for ( int I = 0 ; I <N; I ++) NE [P] [I] = 0 ; 
        CNT [P] = NUM [P] = 0 ; 
        len [P] = L;
        return P ++ ; 
    } 
    void the init () { // Initialization 
        P = 0 ; 
        newNode ( 0 ); 
        newNode ( - . 1 ); // order not anti 
        Last = 0 ; 
        n- = 0 ; 
        S [n-] = - . 1 ; // to prevent cross-border 
        Fail [ 0 ] = . 1 ; 
    } 
    int get_fail ( int X) {
         the while (S [n-- len [X] - . 1!] = S [n-]) = X Fail [X];
         return X; 
    } 
    void the Add ( int C) { 
        C = C - ' A ' ; 
        S [ ++ n-] = C;
         int CUR = get_fail (Last) ; // by a palindromic sequence to find the matching position palindrome string 
        IF (! NE [CUR] [C]) { // If there have been no palindromic sequence, indicate that there is a new different nature back text strings 
            int now newNode = (len [CUR] + 2 ); // New nodes 
            fail [now] NE = [get_fail (fail [CUR])] [C]; // and AC machines as automatic fail to establish a pointer to after the jump mismatch 
            NE [CUR] [C] = now;
            NUM [now]NUM = [Fail [now]] + . 1 ; 
        } 
        Last = NE [CUR] [C]; 
        CNT [Last] ++ ; 
    } 
    void count_cnt () {
         for ( int I = p- . 1 ; I> = 0 ; I - ) { 
            CNT [Fail [I]] + = CNT [I]; // father son accumulated cnt, because if the fail [v] = u, then u must be palindromic sequence sub v! 
        } 
    } 
} PAM; 

char ST [MAXN];
 int main () { 
    Scanf ( " % S " , ST);
     int len = strlen (ST);
    pam.init();
    for(int i = 0; i < len; i++){
       pam.add(st[i]);
    }
    pam.count_cnt();
    ll ma = 0;
    for(int i = 2; i < pam.p; i++){
        ma = max((ll)pam.len[i] * pam.cnt[i], ma);
    }
    cout << ma <<endl;
    return 0;
}