P2414 [NOI2011] Ali typewriter AC automaton

The meaning of problems

Given n pattern string, there are m query, asking each X-th string pattern appears many times in the Y-th mold

Problem-solving ideas

In opposite directions fail to build a tree of tree T, X problem into sub-tree of a termination node number y. Ran dfs sequence of T, X subtrees can be expressed as the period of the interval, and then run again dfs trie, each entry position +1 corresponding to a sequence point on dfs put, put away a point corresponding to the sequence dfs -1 position, traverse to the terminating node y on the calculation results of all inquiry y = y, the summation interval is calculated, segment tree with a tree or array can be easily maintained.

AC Code

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int maxn=2e5+5;
char s[maxn];
struct Query{
    int x,y,id,ans;
    bool operator<(const Query& b)const{return y<b.y;}
}q[maxn];
int ql[maxn],qr[maxn],pos[maxn];



//BIT
int c[maxn];
inline int lowbit(int x){return x&(-x);}
void add(int x,int d){for(;x<maxn-5;x+=lowbit(x))c[x]+=d;}
int getsum(int x){int res=0;for(;x;x-=lowbit(x))res+=c[x];return res;}
int getsum(int l,int r){return getsum(r)-getsum(l-1);}



//AC_Automaton
//root=0,range[1,tot]
const static int SIZE=26; 
int tr[maxn][SIZE],fail[maxn],tot;
int fa[maxn],val[maxn],ch[maxn][SIZE];
void insert(){
    int p=0,cnt=0;
    for(int i=0;s[i];i++){
        if(s[i]=='P'){pos[++cnt]=p;val[p]=cnt;}
        else if(s[i]=='B')p=fa[p];
        else{
            if(!tr[p][s[i]-'a'])tr[p][s[i]-'a']=++tot;
            fa[tr[p][s[i]-'a']]=p;
            p=tr[p][s[i]-'a'];
        }
    }
    for(int i=0;i<=tot;i++)for(int j=0;j<26;j++)ch[i][j]=tr[i][j];
}
void getfail(){
    queue<int>q;
    for(int i=0;i<SIZE;i++)if(tr[0][i])fail[tr[0][i]]=0,q.push(tr[0][i]);
    while(!q.empty()){
        int p=q.front();q.pop();
        for(int i=0;i<SIZE;i++){
            if(tr[p][i]){
                fail[tr[p][i]]=tr[fail[p]][i],q.push(tr[p][i]);
            }
            else tr[p][i]=tr[fail[p]][i];
        }
    }
}




//inv fail tree
struct Edge{
    int v,nxt;
}e[maxn];
int fi[maxn],se[maxn],sz;
int head[maxn],num;
void init_graph(){
    memset(head,0,sizeof(head));
    num=1;
}
void addedge(int u,int v){e[num].v=v;e[num].nxt=head[u];head[u]=num++;}
void dfs_fail(int p){
    fi[p]=++sz;
    for(int i=head[p];i;i=e[i].nxt)dfs_fail(e[i].v);
    se[p]=sz;   
}



//get ans
int Ans[maxn];
void dfs_trie(int p){
    add(fi[p],1);
    if(val[p]){
        for(int i=ql[val[p]];i<=qr[val[p]];i++)
            q[i].ans=getsum(fi[pos[q[i].x]],se[pos[q[i].x]]);
    }
    for(int i=0;i<26;i++)if(ch[p][i])dfs_trie(ch[p][i]);
    add(fi[p],-1);
}

int main()
{
//#ifndef ONLINE_JUDGE
//  freopen("in.txt","r",stdin);
//#endif
    scanf("%s",s);
    insert();
    
    
    getfail();
    init_graph();
    for(int i=1;i<=tot;i++)addedge(fail[i],i);
    sz=0;dfs_fail(0);//建inv fail tree，跑出dfs序 
    
    
    
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d %d",&q[i].x,&q[i].y);
        q[i].id=i;q[i].ans=0;
    }
    sort(q+1,q+1+n);
    for(int i=1,t=1;i<=n;i++){
        ql[q[i].y]=i;
        while(q[t+1].y==q[i].y)t++;
        qr[q[i].y]=t;i=t;
    }
    dfs_trie(0);//遍历trie，跑出答案 
    
    for(int i=1;i<=n;i++)Ans[q[i].id]=q[i].ans;
    for(int i=1;i<=n;i++)printf("%d\n",Ans[i]);
    return 0;
}