【BZOJ3756】Pty的字符串（广义后缀自动机）

题意：

 

 思路：论文题

 

 建立Trie树的后缀自动机需要换这个长的板子

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
typedef pair<ll,int>P;
#define N  800010
#define M  8100000
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const int MOD=1e9+7,inv2=(MOD+1)/2;
      double eps=1e-6;
      int INF=1e9;
      ll inf=5e13;
      int dx[4]={-1,1,0,0};
      int dy[4]={0,0,-1,1};

char s[M];
int p,np,q,nq,k,cas,n;
int pos[N];

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

struct sam
{
    int cnt;
    int fa[N<<1],ch[N<<1][26];
    int st[N<<1],b[N<<1],bl[N<<1],to[N<<1],size[N<<1];
    ll f[N<<1];
    sam()
    {
        cnt=1;
    }

    int add(int p,int x)
    {
        if(ch[p][x])
        {
            q=ch[p][x];
            if(st[q]==st[p]+1)
            {
                size[q]++;
                return q;
            }
             else
             {
                 st[nq=++cnt]=st[p]+1; size[nq]=1;
                 memcpy(ch[nq],ch[q],sizeof ch[q]);
                //t[nq]=t[q];
                 fa[nq]=fa[q];
                 fa[q]=nq;
                 while(ch[p][x]==q)
                 {
                     ch[p][x]=nq;
                     p=fa[p];
                 }
                 return nq;
             }
        }
         else
         {
                st[np=++cnt]=st[p]+1; size[np]=1;
                while(p&&!ch[p][x])
                {
                    ch[p][x]=np;
                    p=fa[p];
                }
                if(!p) fa[np]=1;
                 else
                 {
                        int q=ch[p][x];
                        if(st[q]==st[p]+1) fa[np]=q;
                         else
                         {
                                nq=++cnt; st[nq]=st[p]+1;
                                memcpy(ch[nq],ch[q],sizeof ch[q]);
                                fa[nq]=fa[q];
                                fa[q]=fa[np]=nq;
                                while(ch[p][x]==q)
                                {
                                    ch[p][x]=nq;
                                    p=fa[p];
                                }
                         }
                 }
           }
        return np;
    }

    void solve()
    {
        //printf("cnt=%d\n",cnt);
        rep(i,1,cnt) b[st[i]]++;
        rep(i,1,cnt) b[i]+=b[i-1];
        rep(i,1,cnt) bl[b[st[i]]--]=i;
        scanf("%s",s+1);
        int n=strlen(s+1);
        per(i,cnt,1) size[fa[bl[i]]]+=size[bl[i]];
        //rep(i,1,cnt) printf("%d\n",size[i]);
        size[1]=0;
        rep(i,1,cnt)
        {
            int p=bl[i];
            f[p]=f[fa[p]]+1ll*(st[p]-st[fa[p]])*size[p];
        }
        ll ans=0;
        p=1;
        int L=0;
        rep(i,1,n)
        {
            int x=s[i]-'a';
            if(ch[p][x]) L++,p=ch[p][x];
             else
             {
                 while(p&&!ch[p][x]) p=fa[p];
                 if(p) L=st[p]+1,p=ch[p][x];
                  else L=0,p=1;
             }
             if(p!=1) ans+=f[fa[p]]+1ll*(L-st[fa[p]])*size[p];
        }
        printf("%lld\n",ans);
    }

}sam;

int main()
{
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
    n=read();
    pos[1]=1;
    rep(i,2,n)
    {
        int x=read();
        scanf("%s",s+1);
        pos[i]=sam.add(pos[x],s[1]-'a');
    }
    sam.solve();
    return 0;
}