题意:给一颗字典树,m次查询,每次给出一个字符串,问你该字符串是字典树上多少串的后缀
题解:字典树求广义sam,每次把查询串在sam上跑一遍,最后到达的点的sz就是答案,中途没法走了,就是没有出现过
//#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 1000000007
#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-5;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=1000000+10,maxn=1000000+10,inf=0x3f3f3f3f;
char s[N];
struct SAM{
int last,cnt;
int ch[N<<1][26],fa[N<<1],l[N<<1],sz[N<<1];
int c[N<<1],a[N<<1];
int ins(int y,int x)
{
last=y;
int p=last,np=++cnt;last=np;l[np]=l[p]+1;
for(;p&&!ch[p][x];p=fa[p])ch[p][x]=np;
if(!p)fa[np]=1;
else
{
int q=ch[p][x];
if(l[q]==l[p]+1)fa[np]=q;
else
{
int nq=++cnt;l[nq]=l[p]+1;
memcpy(ch[nq],ch[q],sizeof ch[q]);
fa[nq]=fa[q];fa[q]=fa[np]=nq;
for(;ch[p][x]==q;p=fa[p])ch[p][x]=nq;
}
}
sz[np]=1;
return last;
}
void build()
{
cnt=last=1;
}
void topo()
{
for(int i=1;i<=cnt;i++)c[l[i]]++;
for(int i=1;i<=cnt;i++)c[i]+=c[i-1];
for(int i=1;i<=cnt;i++)a[c[l[i]]--]=i;
}
void gao()
{
topo();
for(int i=cnt;i;i--)sz[fa[a[i]]]+=sz[a[i]];
}
void cal()
{
int len=strlen(s+1),now=1;
for(int i=len;i>=1;i--)
{
if(ch[now][s[i]-'A'])now=ch[now][s[i]-'A'];
else
{
puts("0");
return ;
}
}
printf("%d
",sz[now]);
}
}sam;
int a[N],id[N];
struct Tire{
vector<int>v[N];
void bfs()
{
queue<int>q;q.push(0);id[0]=1;
while(!q.empty())
{
int u=q.front();q.pop();
for(int i=0;i<v[u].size();i++)if(v[u][i])
{
id[v[u][i]]=sam.ins(id[u],a[v[u][i]]);
q.push(v[u][i]);
}
}
}
}t;
int main()
{
int n,m;scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
{
int x;scanf("%s%d",s,&x);
t.v[x].pb(i);a[i]=s[0]-'A';
}
sam.build();
t.bfs();
sam.gao();
while(m--)
{
scanf("%s",s+1);
sam.cal();
}
return 0;
}
/********************
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