CF547E Milk and Friends(AC自动机的fail指针上建主席树 或 广义后缀自动机的parent线段树合并)

What-The-Fatherland is a strange country! All phone numbers there are strings consisting of lowercase English letters. What is double strange that a phone number can be associated with several bears!

In that country there is a rock band called CF consisting of n bears (including Mike) numbered from 1 to n.

Phone number of i-th member of CF is s_i. May 17th is a holiday named Phone Calls day. In the last Phone Calls day, everyone called all the numbers that are substrings of his/her number (one may call some number several times). In particular, everyone called himself (that was really strange country).

Denote as call(i, j) the number of times that i-th member of CF called the j-th member of CF.

The geek Mike has q questions that he wants to ask you. In each question he gives you numbers l, r and k and you should tell him the number

Input

The first line of input contains integers n and q (1 ≤ n ≤ 2 × 10⁵ and 1 ≤ q ≤ 5 × 10⁵).

The next n lines contain the phone numbers, i-th line contains a string s_i consisting of lowercase English letters ().

The next q lines contain the information about the questions, each of them contains integers l, r and k (1 ≤ l ≤ r ≤ n and 1 ≤ k ≤ n).

Output

Print the answer for each question in a separate line.

Examples

Input

5 5
a
ab
abab
ababab
b
1 5 1
3 5 1
1 5 2
1 5 3
1 4 5

Output

7
5
6
3
6

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题意：题目给你N个字符串，Q组询问(1<=N , Q<=200000).(∑s[i] <=200000 )

每组询问，l, r,x  : 在第 l个 字符串  到 第r个字符串这(r-l+1)个字符串中，第x个字符串出现的次数。

题解一（后缀自动机）：

我们将这N个字符串建立SAM，然后将其parent tree拉出来，那么对于一个节点所表示的单词，他出现的次数就是他的子树的大小，由于题目是要求l~r之间，所以是他的子树中endpos在一定范围内的点的数量。我们建立SAM时，对于第k个字符串,每加入一个节点，就在对应节点的线段树里面的第k个位置加一。 我们用线段树维护每个节点的子树，然后线段树合并。

对于每一个询问，我们只要在对应的子树里面找l~r区间的和就行了。

参考代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int,int>
#define pil pair<int,ll>
#define mkp make_apir
const int INF=0x3f3f3f3f;
const int maxn=2e5+;
int N,Q;
char s[maxn];

int nxt[maxn<<][],l[maxn<<],fa[maxn<<];
int last,tot,cnt[maxn<<],c[maxn<<];
int sz,p[maxn],T[maxn<<];

struct Tr{
    int ls,rs;
    int num;
} tr[maxn*];

void Update(int &x,int l,int r,int pos)
{
    if(!x) x=++sz;
    tr[x].num++;
    if(l==r) return ;
    int mid=l+r>>;
    if(pos<=mid) Update(tr[x].ls,l,mid,pos);
    else Update(tr[x].rs,mid+,r,pos);
}

int Query(int x,int l,int r,int L,int R)
{
    if(!x) return ;
    if(L<=l&&r<=R) return tr[x].num;
    int mid=l+r>>,res=;
    if(L<=mid) res+=Query(tr[x].ls,l,mid,L,R);
    if(R>mid) res+=Query(tr[x].rs,mid+,r,L,R);
    return res;
}

int Merge(int x,int y)
{
    if(!x||!y) return x+y;
    int z=++sz;
    tr[z].ls=Merge(tr[x].ls,tr[y].ls);
    tr[z].rs=Merge(tr[x].rs,tr[y].rs);
    tr[z].num=tr[x].num+tr[y].num;
    return z;
}

void Mer()
{
    for(int i=;i<=tot;++i) cnt[l[i]]++;
    for(int i=;i<=tot;++i) cnt[i]+=cnt[i-];
    for(int i=;i<=tot;++i) c[cnt[l[i]]--]=i;
    for(int i=tot,x;i>;--i) x=c[i],T[fa[x]]=Merge(T[x],T[fa[x]]);
}

void Init()
{
    last=tot=; sz=;
    memset(nxt[tot],,sizeof nxt[tot]);
    l[tot]=fa[tot]=;
}

int NewNode()
{
    ++tot;
    memset(nxt[tot],,sizeof nxt[tot]);
    l[tot]=fa[tot]=;
    return tot;
}

void Insert(int ch,int x)
{
    int p,q,np,nq;
    if(nxt[last][ch])
    {
        p=last;q=nxt[p][ch];
        if(l[q]==l[p]+) last=q;
        else
        {
            nq=NewNode();
            l[nq]=l[p]+;
            memcpy(nxt[nq],nxt[q],sizeof(nxt[q]));
            fa[nq]=fa[q];fa[q]=nq;
            while(p&&nxt[p][ch]==q) nxt[p][ch]=nq,p=fa[p];
            last=nq;
        }
    }
    else
    {
        np=NewNode(),p=last;
        last=np; l[np]=l[p]+;
        while(p&&!nxt[p][ch]) nxt[p][ch]=np,p=fa[p];
        if(!p) fa[np]=;
        else
        {
            q=nxt[p][ch];
            if(l[q]==l[p]+) fa[np]=q;
            else
            {
                nq=NewNode();
                memcpy(nxt[nq],nxt[q],sizeof nxt[q]);
                fa[nq]=fa[q];
                l[nq]=l[p]+;
                fa[q]=fa[np]=nq;
                while(p&&nxt[p][ch]==q) nxt[p][ch]=nq,p=fa[p];
            }
        }
    }
}

int main()
{
    scanf("%d%d",&N,&Q);
    Init();
    for(int i=;i<=N;++i)
    {
        scanf("%s",s);
        int len=strlen(s); last=;
        for(int j=;j<len;++j) Insert(s[j]-'a',i),Update(T[last],,N,i);
        p[i]=last;
    }
    Mer();

    while(Q--)
    {
        int l,r,x;
        scanf("%d%d%d",&l,&r,&x);
        printf("%d\n",Query(T[p[x]],,N,l,r));
    }

    return ;
}

题解二（AC自动机）：首先对于多串匹配问题，我们很容易想到AC自动机。在AC自动机的fail树中的每个节点,我们知道他出现的次数就是他的子树的大小。但是这题是要询问在第 l个 字符串  到 第r个字符串这(r-l+1)个字符串中，第x个字符串出现的次数，这我们可以想到主席树的询问历史版本信息的性质，那么我们就可以用主席树去维护AC自动机的fail指针的dfs序。然后我们只要求出第R棵树和地(L-1)棵树的差值就行了。

参考代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int,int>
#define pil pair<int,ll>
#define mkp make_apir
const int INF=0x3f3f3f3f;
const int maxn=2e5+;
int N,Q;
char s[maxn];
int sz,ch[maxn][],fail[maxn],val[maxn];
int in[maxn],out[maxn],tim,Len[maxn],tot,T[maxn];
vector<int> vec[maxn],ts[maxn];
void Init()
{
    sz=; tim=tot=;
    memset(ch[sz],,sizeof ch[sz]);
    memset(val,,sizeof val);
}
int idx(char c){return c-'a';}
void Insert(char *s,int x)
{
    Len[x]=strlen(s);int u=;
    for(int i=;i<Len[x];++i)
    {
        int c=idx(s[i]); ts[x].push_back(c);
        if(!ch[u][c]){memset(ch[sz],,sizeof ch[sz]);ch[u][c]=sz++;}
        u=ch[u][c];
    }
    val[x]=u;
}
void GetFail()
{
    queue<int> q;
    fail[]=;
    for(int i=;i<;++i)
        if(ch[][i]) {fail[ch[][i]]=;q.push(ch[][i]);}

    while(!q.empty())
    {
        int u=q.front();q.pop();
        vec[fail[u]].push_back(u);
        for(int i=;i<;++i)
        {
            int c=ch[u][i];
            if(!c){ch[u][i]=ch[fail[u]][i];continue;}
            q.push(c);
            fail[c]=ch[fail[u]][i];
        }
    }
}
void dfs(int u)
{
    in[u]=++tim;
    for(int i=,len=vec[u].size();i<len;++i)
        dfs(vec[u][i]);
    out[u]=tim;
}

struct Tree{
    int ls,rs;
    int num;
} tr[maxn*];

inline void Update(int y,int &x,int l,int r,int pos)
{
    tr[++tot]=tr[y];tr[tot].num++; x=tot;
    if(l==r) return;
    int mid=l+r>>;
    if(pos<=mid) Update(tr[y].ls,tr[x].ls,l,mid,pos);
    else Update(tr[y].rs,tr[x].rs,mid+,r,pos);
}
inline void Build(int y,int &x,int id)
{
    int now=,last=y;
    for(int i=;i<Len[id];++i)
    {
        now=ch[now][ts[id][i]];
        Update(last,x,,tim,in[now]);
        last=x;
    }
}
inline int Query(int y,int x,int l,int r,int L,int R)
{
    if(L<=l&&r<=R) return tr[x].num-tr[y].num;
    int mid=l+r>>,res=;
    if(L<=mid) res+=Query(tr[y].ls,tr[x].ls,l,mid,L,R);
    if(R>mid) res+=Query(tr[y].rs,tr[x].rs,mid+,r,L,R);
    return res;
}

int main()
{
    scanf("%d%d",&N,&Q);
    Init();
    for(int i=;i<=N;++i) scanf("%s",s),Insert(s,i);
    GetFail();
    dfs();
    for(int i=;i<=N;++i) Build(T[i-],T[i],i);
    while(Q--)
    {
        int l,r,x;
        scanf("%d%d%d",&l,&r,&x);
        printf("%d\n",Query(T[l-],T[r],,tim,in[val[x]],out[val[x]]));
    }

    return ;
}