回文自动机pam

目的：类似回文Trie树+ac自动机，可以用来统计一些其他的回文串相关的量

复杂度：O(nlogn)

https://blog.csdn.net/Lolierl/article/details/99971257

https://www.luogu.org/problem/P5496

求出以每个位置结尾的回文子串个数，强制在线

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=2e6+;
struct pam_trie
{
    int ch[];
    int fail,len,num;
};
struct pam
{
    pam_trie b[maxn];
    int n,length,last,cnt,s[maxn];
    char c[maxn];
    pam()
    {
        b[].len=;b[].len=-;
        b[].fail=;b[].fail=;
        last=;
        cnt=;
    }
    void read()
    {
        scanf("%s",c+);
        length=strlen(c+);
    }
    int get_fail(int x)
    {
        while(s[n-b[x].len-]!=s[n])x=b[x].fail;
        return x;
    }
    void insert()
    {
        int p=get_fail(last);
        if(!b[p].ch[s[n]])
        {
            b[++cnt].len=b[p].len+;
            b[cnt].fail=b[get_fail(b[p].fail)].ch[s[n]];
            //b[cnt].num=b[b[cnt].fail].num+1;
            b[p].ch[s[n]]=cnt;
        }
        last=b[p].ch[s[n]];
        b[last].num=b[b[last].fail].num+;
    }
    void solve()
    {
        int k=;
        s[]=;
        for(n=;n<=length;n++)
        {
            c[n]=(c[n]-+k)%+;
            s[n]=c[n]-'a';
            insert();
            printf("%d ",b[last].num);
            k=b[last].num;
        }
    }
}P;

int main()
{
    P.read();
    P.solve();
    return ;
}

https://www.luogu.org/problem/P3649

求回文子串出现次数*长度的最大值

#include<iostream>
#include<stdio.h>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=3e5+;
struct pam_trie
{
    int ch[];
    int fail,len,sum;
};
struct pam
{
    pam_trie b[maxn];
    int n,length,last,cnt,s[maxn];
    char c[maxn];
    long long ans;
    pam()
    {
        b[].len=;b[].len=-;
        b[].fail=;b[].fail=;
        last=;
        cnt=;
    }
    void read()
    {
        scanf("%s",c+);
        length=strlen(c+);
    }
    int get_fail(int x)
    {
        while(s[n-b[x].len-]!=s[n])x=b[x].fail;
        return x;
    }
    void insert()
    {
        int p=get_fail(last);
        if(!b[p].ch[s[n]])
        {
            b[++cnt].len=b[p].len+;
            b[cnt].fail=b[get_fail(b[p].fail)].ch[s[n]];
            b[p].ch[s[n]]=cnt;
        }
        last=b[p].ch[s[n]];
        b[last].sum++;
    }
    void solve()
    {
        s[]=;
        for(n=;n<=length;n++)
        {
            s[n]=c[n]-'a';
            insert();
        }
        ans=;
        for(int i=cnt;i>;i--)
        {
            b[b[i].fail].sum+=b[i].sum;
            ans=max(ans,1ll*b[i].sum*b[i].len);
        }
        printf("%lld\n",ans);
    }
}P;
int main()
{
    P.read();
    P.solve();
}

https://www.luogu.org/problem/P4287

计算串的最长双倍回文子串的长度，tips：fail指针指向当前节点所表示的回文串的最长回文后缀

#include<stdio.h>
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=5e5+;
struct pam_trie
{
    int ch[];
    int fail,len,sum;
};
struct pam
{
    pam_trie b[maxn];
    int n,length,last,cnt,s[maxn];
    char c[maxn];
    pam()
    {
        b[].len=;b[].len=-;
        b[].fail=;b[].fail=;
        last=;cnt=;
    }
    void read()
    {
        scanf("%d",&length);
        scanf("%s",c+);
    }
    int get_fail(int x)
    {
        while(s[n-b[x].len-]!=s[n])x=b[x].fail;
        return x;
    }
    void insert()
    {
        int p=get_fail(last);
        if(!b[p].ch[s[n]])
        {
            b[++cnt].len=b[p].len+;
            b[cnt].fail=b[get_fail(b[p].fail)].ch[s[n]];
            b[p].ch[s[n]]=cnt;
        }
        last=b[p].ch[s[n]];
        b[last].sum++;
    }
    void solve()
    {
        s[]=;
        for(n=;n<=length;n++)
        {
            s[n]=c[n]-'a';
            insert();
        }
        int ans=;
        for(int i=cnt;i>;i--)
        {
            int pos=i;
            if(b[i].len%!=||b[i].len<=ans)continue;
            while(*b[pos].len>b[i].len)pos=b[pos].fail;
            if(*b[pos].len==b[i].len)ans=b[i].len;
        }
        printf("%d\n",ans);
    }
}P;
int main()
{
    P.read();
    P.solve();
    return ;
}

ICPC 2018 南京 Mediocre String Problem，用回文自动机来求出以每个位置结尾的回文子串个数，再进行exkmp

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=1e6+;
struct pam_trie
{
    int ch[];
    int fail,len,num;
};
int res[maxn];
char s1[maxn],s2[maxn],t[maxn];

struct pam
{
    pam_trie b[maxn];
    int n,last,cnt,s[maxn],length;
    pam()
    {
        b[].len=;b[].len=-;
        b[].fail=;b[].fail=;
        last=;
        cnt=;
    }
    int get_fail(int x)
    {
        while(s[n-b[x].len-]!=s[n])x=b[x].fail;
        return x;
    }
    int insert()
    {
        int p=get_fail(last);
        if(!b[p].ch[s[n]])
        {
            b[++cnt].len=b[p].len+;
            b[cnt].fail=b[get_fail(b[p].fail)].ch[s[n]];
            b[p].ch[s[n]]=cnt;
        }
        last=b[p].ch[s[n]];
        b[last].num=b[b[last].fail].num+;
        return b[last].num;
    }
    void solve()
    {
        s[]=;
        length=strlen(s1);
        for(n=;n<=length;n++)
        {
            s[n]=s1[n-]-'a';
            res[n-]=insert();
        }
    }
}P;

int Next[maxn],extend[maxn];
void get_next(char *s)
{
    int n=strlen(s),i,j,k=;
    for(j=;+j<n&&s[j]==s[+j];j++);
    Next[]=j;
    for(i=;i<n;i++)
    {
        int len=k+Next[k],L=Next[i-k];
        if(L<len-i)Next[i]=L;
        else
        {
            for(j=max(,len-i);i+j<n&&s[j]==s[i+j];j++);
            Next[i]=j;
            k=i;
        }
    }
    Next[]=n;
}
void ex_kmp(char *T,char *s)
{
    int n=strlen(T),m=strlen(s),i,j,k;
    for(j=;j<n&&j<m&&T[j]==s[j];j++);
    extend[]=j;
    k=;
    for(i=;i<n;i++)
    {
        int len=k+extend[k],L=Next[i-k];
        if(L<len-i)extend[i]=L;
        else
        {
            for(j=max(,len-i);j<m&&i+j<n&&s[j]==T[i+j];j++);
            extend[i]=j;
            k=i;
        }
    }
}

int main()
{
    scanf("%s",s1);
    scanf("%s",t);
    int lens=strlen(s1);

    reverse(s1,s1+lens);
    for(int i=;i<lens;i++)s2[i]=s1[i];
    s2[lens]='\0';

    get_next(t);
    ex_kmp(s2,t);
    long long ans=;
    P.solve();
    for(int i=;i<lens;i++)
    {
        ans+=1ll*extend[i]*res[i-];
    }
    printf("%lld\n",ans);
    return ;
} 

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