【Manacher】Colorful String

The value of a string s is equal to the number of different letters which appear in this string.

Your task is to calculate the total value of all the palindrome substring.

Input
The input consists of a single string |s|(≤∣s∣≤×^ ).

The string s only contains lowercase letters.

Output
Output an integer that denotes the answer.

样例输入
abac

样例输出 

样例解释
abac has palindrome substrings a,b,a,c,abaa,b,a,c,aba，ans the total value is equal to ++++=。

【题解】

　　Manacher，先预处理一波，然后找出每一个位置的26个字母下一个位置，存在一个vector里面，然后最后找的时候就是差值 × 对应的个数。

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const int N=3e5+;
 char S[N],T[N<<];
 int Len[N*];
 int nxt[N][];
 vector<int>V[N];
 void Manacher(char *s){
     int L=strlen(s);
     int po =  , mx= ;

     for(int i=;i<=L*;i++){
         T[i] = i&? '#' : s[i/-] ;
     }

     T[] = '@';
     T[*L+] = '#';
     T[*L+] = '\0';

     ll tmp =  ;
     for(int i=;i<=*L;i++){
         if( i<mx ){
             Len[i]=min( mx-i , Len[*po-i] );
         }else{
             Len[i] =  ;
         }

         while( T[i+Len[i]]==T[i-Len[i]] ){
             Len[i]++;
         }
         if( i + Len[i] > mx ){
             po = i;
             mx = i + Len[i];
         }
     }
 }
 int main()
 {
     scanf("%s",S);
     Manacher(S);
     int len1 = strlen( S ) ;
     int len2 = strlen( T ) ;

     for( int j =  ; j <  ; j++ ){
         int pos = INT_MAX ;
         for( int i = len1 -  ; i >=  ; i-- ){
             if( S[i] == j + 'a' ) pos = i ;
             nxt[i][j] = pos ;
         }
     }

     for( int i =  ; i < len1 ; i++ ){
         for( int j =  ; j <  ; j++ ){
             V[i].push_back( nxt[i][j] ) ;
         }
         sort( V[i].begin() , V[i].end() );
     }
     /*
     for( int i = 1 ; i < len2 ; i++ ){
         printf("%d : %c , Len %d \n",i , T[i] , Len[i] );
     }
     */
     ll ans =  ;
     for( int i =  ; i < len2 ; i++ ){
         int L = Len[i] /  ;
         if( L ==  ) continue ;

         int Id = i /  -  ;
         if( i &  ) Id ++ ;

         int RR = Id + L -  ;
         int Last = V[Id][] ;
         int cnt =   ;
         for( auto x : V[Id] ){
             if( x > RR ) break ;
             int r = x -  ;
             ans = ans + (ll)( cnt ++ ) * (ll) ( r - Last +  );
             Last = x ;
         }
         if( RR >= Last ){
             ans = ans + (ll) cnt * (ll) ( RR - Last +  );
         }
     }
     printf("%lld\n",ans);
     return ;
 }

 //jswbutgecnmqnuagqnvxfljfffzvudcfvygpro