hdu5421Victor and String 两端加点的pam

时间：2023-08-09 12:36:02

题意:要求维护两端加点的字符串,以及查询本质回文串个数和所有回文串个数

题解:pam,两端加点过程详见ioi2017国家集训队论文,维护一个最长回文前缀和最长回文后缀即可,fail不用两个,能前后共用一个.维护所有回文串个数,就是用增量法,每加一个点计算含这个点的回文串的个数(即当前回文后缀),即fail链到0或1的长度,维护一个dep表示fail链长度即可.

//#pragma GCC optimize(2)

//#pragma GCC optimize(3)

//#pragma GCC optimize(4)

//#pragma GCC optimize("unroll-loops")

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include<bits/stdc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 998244353

#define ld long double

//#define C 0.5772156649

//#define ls l,m,rt<<1

//#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#define pil pair<int,ll>

#define pli pair<ll,int>

#define pii pair<int,int>

#define ull unsigned long long

//#define base 1000000000000000000

#define fin freopen("a.txt","r",stdin)

#define fout freopen("a.txt","w",stdout)

#define fio ios::sync_with_stdio(false);cin.tie(0)

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;

const db eps=1e-7;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=200000+10,maxn=100000+10,inf=0x3f3f3f3f;

struct PAM{

    int ch[N][26],fail[N],len[N],s[N],dep[N];

    int last[2],l,r,p;

    ll ans;

    int newnode(int w)

    {

        for(int i=0;i<26;i++)ch[p][i] = 0;

        len[p] = w;

        dep[p] = 0;

        return p++;

    }

    void init()

    {

        l = 100000+1,r = 100000;

        p = ans = 0;

        last[0]=last[1]=0;

        newnode(0);

        newnode(-1);

        memset(s,-1,sizeof s);

        fail[0] = 1;

    }

    int getfail(int x,int op)

    {

        if(op)while(s[r-len[x]-1] != s[r]) x = fail[x];

        else while(s[l+len[x]+1] != s[l]) x = fail[x];

        return x;

    }

    void add(int c,int op)

    {

        if(op)s[++r] = c;

        else s[--l] = c;

        int cur = getfail(last[op],op);

        if(!ch[cur][c]){

            int now = newnode(len[cur]+2);

            fail[now] = ch[getfail(fail[cur],op)][c];

            dep[now] = dep[fail[now]] + 1;

            ch[cur][c] = now;

        }

        last[op] = ch[cur][c];

//        printf("%d %d %d %d\n",op,cur,last[op],fail[last[op]]);

        if(len[last[op]] == r-l+1)last[op^1]=last[op];

        ans+=dep[last[op]];

    }

}pam;

char s[10];

int main()

{

    int n;

    while(~scanf("%d",&n))

    {

        pam.init();

        for(int i=1;i<=n;i++)

        {

            int x;

            scanf("%d",&x);

            if(x<=2)

            {

                scanf("%s",s);

                pam.add(s[0]-'a',x-1);

            }

            else if(x==3)printf("%d\n",pam.p-2);

            else printf("%lld\n",pam.ans);

        }

    }

    return 0;

}

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