CF G. Indie Album AC自动机+fail树+线段树

这个套路挺有意思的. 

把 $trie$ 和 $fail$ 树都建出来，然后一起跑一跑就好了~

#include <queue> 
#include <cstdio>  
#include <vector> 
#include <cstring> 
#include <algorithm>
#define N 500004 
#define setIO(s) freopen(s".in","r",stdin) 
using namespace std;  
struct Seg {    
	#define lson (now<<1) 
	#define rson (now<<1|1) 
	int sum[N<<2]; 
	void update(int l,int r,int now,int p,int v) {
		sum[now]+=v; 
		if(l==r) return; 
		int mid=(l+r)>>1; 
		if(p<=mid) update(l,mid,lson,p,v); 
		else update(mid+1,r,rson,p,v);  
	} 
	int query(int l,int r,int now,int L,int R) {
		if(l>=L&&r<=R) return sum[now]; 
		int mid=(l+r)>>1,re=0; 
		if(L<=mid) re+=query(l,mid,lson,L,R); 
		if(R>mid) re+=query(mid+1,r,rson,L,R); 
		return re; 
	}
}seg; 
struct Trie {
	int c,to;  
	Trie(int c=0,int to=0):c(c),to(to){}   
};   
struct Node {
	int ch[27]; 
}t[N];       
struct AC {
	int ch[27],f; 
}a[N];      
struct Q {
	int i,id;       
	Q(int i=0,int id=0):i(i),id(id){} 
}; 
queue<int>q; 
vector<Q>G[N];  
vector<Trie>T[N];        
char S[N]; 
int n,tot,m,edges,tim,siz[N],tot2,id[N],endd[N],hd[N],nex[N],to[N],dfn[N],answer[N];      
inline void add(int u,int v) {
	nex[++edges]=hd[u],hd[u]=edges,to[edges]=v; 
}
inline void insert(int x) {  
	int rt=0,len=strlen(S+1),i; 
	for(i=1;i<=len;++i) {
		if(!a[rt].ch[S[i]-'a']) a[rt].ch[S[i]-'a']=++tot2;   
		rt=a[rt].ch[S[i]-'a'];  
	} 
	endd[x]=rt;      
} 
inline void buildAC() {
	for(int i=0;i<27;++i) if(a[0].ch[i]) q.push(a[0].ch[i]);  
	while(!q.empty()) {
		int u=q.front(),i; 
		q.pop(); 
		for(i=0;i<27;++i) {
			int p=a[u].ch[i]; 
			if(!p) {
				a[u].ch[i]=a[a[u].f].ch[i];   
				continue; 
			}
			a[p].f=a[a[u].f].ch[i];          
			q.push(p);            
		}
	}  
}
void dfs(int u) {
	siz[u]=1,dfn[u]=++tim; 
	for(int i=hd[u];i;i=nex[i]) {
		dfs(to[i]),siz[u]+=siz[to[i]]; 
	}  
}
inline void build_tree() {
	int i,j; 
	for(i=1;i<=tot2;++i) add(a[i].f,i);    
	dfs(0);   
}
void solve(int now,int x) {       
	seg.update(1,tim,1,dfn[now],1);              
	for(int i=0;i<G[x].size();++i) {
		answer[G[x][i].i]=seg.query(1,tim,1,dfn[G[x][i].id],dfn[G[x][i].id]+siz[G[x][i].id]-1);  
	}
	for(int i=0;i<T[x].size();++i) {
		solve(a[now].ch[T[x][i].c],T[x][i].to); 
	} 
	seg.update(1,tim,1,dfn[now],-1);                         
}
int main() {
	int i,j; 
	// setIO("input"); 
	scanf("%d",&n);  
	for(i=1;i<=n;++i) {
		int op,x,lst=0;
		char str[2];   
		scanf("%d",&op); 
		if(op==2) scanf("%d",&lst),lst=id[lst];   
		scanf("%s",str);  
		if(!t[lst].ch[str[0]-'a']) {        
			t[lst].ch[str[0]-'a']=++tot; 
			id[i]=t[lst].ch[str[0]-'a']; 
		    T[lst].push_back(Trie(str[0]-'a',id[i]));             
		}      
		else id[i]=t[lst].ch[str[0]-'a']; 
 	}   
 	scanf("%d",&m); 
 	for(i=1;i<=m;++i) {  
 		scanf("%d%s",&j,S+1),insert(i),G[id[j]].push_back(Q(i, endd[i]));           
 	}   
 	buildAC();
 	build_tree();   
 	solve(0,0);    
 	for(i=1;i<=m;++i) printf("%d
",answer[i]);  
	return 0; 
}