算法学习：回文自动机

Posted 2022-06-23 rentu

【定义】

【自动机】

参照AC自动机

 

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【前置知识】

【AC自动机】

【manacher】

其实不学这两个也可以，但是学过之后会更方便理解

 

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【解决问题】

主要解决回文串的问题

能求出   字符串中回文子串的长度和出现次数

 

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#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#define ll long long
using namespace std;
const int mod = 19930726;
const int MAXN = 1000010;
ll k;
int n, cnt;
int now;
struct note

    int son[26];
    ll len;
    int  siz;
trie[MAXN];
int fail[MAXN], pos;
bool operator<(note a, note b)

    return a.len > b.len;

char s[MAXN];
int get(int x)

    while (s[pos] != s[pos - trie[x].len - 1]) x = fail[x];
    return x;

void insert(int x)

    int cur = get(now);
    if (!trie[cur].son[x])
    
        int u = ++cnt;
        trie[u].len = trie[cur].len + 2;
        fail[u] = trie[get(fail[cur])].son[x];
        trie[cur].son[x] = u;
    
    trie[trie[cur].son[x]].siz++;
    now = trie[cur].son[x];

ll poww(ll a, int b)

    ll res = 1;
    while (b)
    
        if (b & 1)
        
            res = (res*a) % mod;
        
        a = (a*a) % mod;
        b = b >> 1;
    
    return res;

void init()

    fail[0] = 1; fail[1] = 0;
    now = 0, cnt = 1;
    trie[1].len = -1;

int main()

    scanf("%d%lld", &n, &k);
    scanf("%s", s+1);
    init();
    s[0] = 0;
    for (pos = 1; pos <= n; pos++)
        insert(s[pos] - ‘a‘);

    for (int i = cnt; i >= 2; i--)
        trie[fail[i]].siz = (trie[fail[i]].siz + trie[i].siz) % mod;
    //for (int i = 1; i <= cnt; i++)
        //printf("%lld %d\\n", trie[i].len, trie[i].siz);
    sort(trie + 1, trie + 1 + cnt);
    ll ans = 1;
    int pos = 1;
    while (k)
    
        if (pos > cnt)
        
            printf("-1");
            return 0;
        
        if (trie[pos].len % 2 == 0)
        
            pos++;
            continue;
        
        ans = (ans*poww(trie[pos].len, k < trie[pos].siz ? k : trie[pos].siz)) % mod;
        k -= k < trie[pos].siz ? k : trie[pos].siz;
        pos++;
    
    printf("%lld", ans);
    return 0;

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