【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|(1≤∣s∣≤3×10^5 ).

The string s only contains lowercase letters.

Output
Output an integer that denotes the answer.

样例输入 
abac

样例输出 
6

样例解释
abac has palindrome substrings a,b,a,c,abaa,b,a,c,aba，ans the total value is equal to 1+1+1+1+2=6。

【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+100;
char S[N],T[N<<1];
int Len[N*2];
int nxt[N][27];
vector<int>V[N];
void Manacher(char *s){
    int L=strlen(s);
    int po = 0 , mx=0 ;

    for(int i=1;i<=L*2;i++){
        T[i] = i&1? '#' : s[i/2-1] ;
    }

    T[0] = '@';
    T[2*L+1] = '#';
    T[2*L+2] = '';

    ll tmp = 0 ;
    for(int i=1;i<=2*L;i++){
        if( i<mx ){
            Len[i]=min( mx-i , Len[2*po-i] );
        }else{
            Len[i] = 1 ;
        }

        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 = 0 ; j < 26 ; j++ ){
        int pos = INT_MAX ;
        for( int i = len1 - 1 ; i >= 0 ; i-- ){
            if( S[i] == j + 'a' ) pos = i ;
            nxt[i][j] = pos ;
        }
    }


    for( int i = 0 ; i < len1 ; i++ ){
        for( int j = 0 ; j < 26 ; 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 
",i , T[i] , Len[i] );
    }
    */
    ll ans = 0 ;
    for( int i = 2 ; i < len2 ; i++ ){
        int L = Len[i] / 2 ;
        if( L == 0 ) continue ;

        int Id = i / 2 - 1 ;
        if( i & 1 ) Id ++ ;

        int RR = Id + L - 1 ;
        int Last = V[Id][0] ;
        int cnt = 0  ;
        for( auto x : V[Id] ){
            if( x > RR ) break ;
            int r = x - 1 ;
            ans = ans + (ll)( cnt ++ ) * (ll) ( r - Last + 1 );
            Last = x ;
        }
        if( RR >= Last ){
            ans = ans + (ll) cnt * (ll) ( RR - Last + 1 );
        }
    }
    printf("%lld
",ans);
    return 0;
}

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