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

最新推荐文章于 2025-05-11 11:03:35 发布

Top_Spirit

最新推荐文章于 2025-05-11 11:03:35 发布

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分类专栏：回文自动机文章标签： acm 字符串回文自动机

本文链接：https://blog.youkuaiyun.com/weixin_41190227/article/details/87274454

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本文介绍了一种使用回文自动机解决动态增加字符时如何快速计算当前字符串中不同回文子串数量的方法。通过构建回文树并在过程中输出每个前缀对应的回文子串数量，实现了高效解题。

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

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 <bits/stdc++.h>
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 ;
}