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 si. 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 si 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
5 5
a
ab
abab
ababab
b
1 5 1
3 5 1
1 5 2
1 5 3
1 4 5
7
5
6
3
6
题意：题目给你N个字符串，Q组询问(1<=N , Q<=200000).(∑s[i] <=200000 )
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 ;
}