Palindromes and Super Abilities URAL - 1960 （回文自动机）

After solving seven problems on Timus Online Judge with a word “palindrome” in the problem name, Misha has got an unusual ability. Now, when he reads a word, he can mentally count the number of unique nonempty substrings of this word that are palindromes.

Dima wants to test Misha’s new ability. He adds letters s 1, ..., s n to a word, letter by letter, and after every letter asks Misha, how many different nonempty palindromes current word contains as substrings. Which n numbers will Misha say, if he will never be wrong?

Input

The only line of input contains the string s 1... s n, where s i are small English letters (1 ≤ n ≤ 10 5).

Output

Output n numbers separated by whitespaces, i-th of these numbers must be the number of different nonempty substrings of prefix s 1... s i that are palindromes.

Example

input	output
aba	1 2 3

题目大意：输出每个前缀本质不同的回文子串的个数。

解题思路：回文自动机刚开始有两个结点，之后每次多出一个结点，就说明有几个不同的回文子串，所以在一边建树一边输出就可以了，不要忘了减去最开始那两个结点。

/*
@Author: Top_Spirit
@Language: C++
*/
#include 
using namespace std ;
typedef unsigned long long ull ;
typedef long long ll ;
const int Maxn = 1e5 + 10 ;
const int INF = 0x3f3f3f3f ;
const double PI = acos(-1.0) ;
const ull seed = 133 ;

int n ;

struct palindromic_tree{
    int Next[Maxn][26] ;
    int fail[Maxn] ;
    int cnt[Maxn] ;
    int num[Maxn] ;
    int len[Maxn] ;
    int s[Maxn] ;
    int last ;
    int n, p ;
    int newNode (int k){
        for (int i = 0; i < 26; i++) Next[p][i] = 0 ;
        cnt[p] = 0 ;
        num[p] = 0 ;
        len[p] = k ;
        return p++ ;
    }
    void init(){
        p = 0 ;
        newNode(0) ;
        newNode(-1) ;
        last = 0 ;
        n = 0 ;
        s[n] = -1 ;
        fail[0] = 1 ;
    }
    int get_fail (int x){
        while (s[n - len[x] - 1] != s[n]) x = fail[x] ;
        return x ;
    }
    void add (int c){
        c -= 'a' ;
        s[++n] = c ;
        int cur = get_fail(last) ;
        if (!Next[cur][c]){
            int Now = newNode(len[cur] + 2) ;
            fail[Now] = Next[get_fail(fail[cur])][c] ;
            Next[cur][c] = Now ;
            num[Now] = num[fail[Now]] + 1 ;
        }
        last = Next[cur][c] ;
        cnt[last]++ ;
    }
    void Count(){
        for (int i = p - 1; i >= 0; i--){
            cnt[fail[i]] += cnt[i] ;
        }
    }
}Tree;

char str[Maxn] ;

int main (){
    cin >> str ;
    n = strlen(str) ;
    Tree.init() ;
    for (int i = 0; i < n; i++){
        Tree.add(str[i]) ;
        cout << Tree.p - 2 ;
        if (i < n - 1) cout << " " ;
    }
    cout << endl ;
    return 0 ;
}