Consider l = 1, r = | s | divided portion, s t is not necessary to calculate each sub-string of the longest prefix of suffix, referred to as pp [k], has the following limitations:
1.pp [POS [K ]] <len (pos [k ] represents a certain end position k), because they can not be matched
1.pp [POS [K ]] <len (pos [k ] represents a certain end position k), because they can not be matched
2.len [fa [k]] <len <= len [k], since the strings have been only this point
Thereby obtaining the answer is sigma (max (0, len [j] -max (len [fa [j]], pp [pos [j]]))) (0 To take max)
Maintaining a suffix automaton from li s ri to, but does not seem easy. You may wish to change the angle, i.e., the right set of requirements exist which [li, ri] between a value
That is also calculated in the SAM s, the right to obtain a set of nodes each
Maintenance segment tree, the tree line mark indicates the position, value indicating whether the collection is right. Then his father on the set SAM is the right of all sons and, that is, the segment tree merge
(Note: Because you want to keep involved in the merger of the tree line, so to create a new node)
While the segment tree can easily judge there interval 1, and to find the interval
Note, however, can not directly jump FA at this time, but a reduction to a length (as it may not match the longest string of the node, but it can match the shorter string)
View Code
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define N 2000005 4 #define mid (l+r>>1) 5 struct ji{ 6 int nex,to; 7 }edge[N]; 8 int V,x,last,fa[N],pos[N],len[N],ch[N][26]; 9 int VV,E,n,m,a,b,r[N],head[N],pp[N],ls[N*16],rs[N*16]; 10 long long ans; 11 char s[N],t[N]; 12 int New(){ 13 fa[++V]=0; 14 len[V]=pos[V]=0; 15 memset(ch[V],0,sizeof(ch[V])); 16 return V; 17 } 18 void add_edge(int x,int y){ 19 edge[E].nex=head[x]; 20 edge[E].to=y; 21 head[x]=E++; 22 } 23 void add(int c,int id){ 24 int p=last,np=last=New(); 25 len[np]=len[p]+1; 26 pos[V]=id; 27 for(;(p)&&(!ch[p][c]);p=fa[p])ch[p][c]=np; 28 if (!p)fa[np]=x; 29 else{ 30 int q=ch[p][c]; 31 if (len[q]==len[p]+1)fa[np]=q; 32 else{ 33 int nq=New(); 34 pos[nq]=pos[q]; 35 len[nq]=len[p]+1; 36 memcpy(ch[nq],ch[q],sizeof(ch[q])); 37 fa[nq]=fa[q]; 38 fa[q]=fa[np]=nq; 39 for(;(p)&&(ch[p][c]==q);p=fa[p])ch[p][c]=nq; 40 } 41 } 42 } 43 void update(int &k,int l,int r,int x){ 44 k=++VV; 45 if (l==r)return; 46 if (x<=mid)update(ls[k],l,mid,x); 47 else update(rs[k],mid+1,r,x); 48 } 49 int query(int k,int l,int r,int x,int y){ 50 if ((!k)||(l>y)||(x>r))return 0; 51 if ((x<=l)&&(r<=y))return 1; 52 return query(ls[k],l,mid,x,y)|query(rs[k],mid+1,r,x,y); 53 } 54 void merge(int &k1,int k2){ 55 if (k1*k2==0){ 56 k1+=k2; 57 return; 58 } 59 ls[++VV]=ls[k1]; 60 rs[VV]=rs[k1]; 61 k1=VV; 62 merge(ls[k1],ls[k2]); 63 merge(rs[k1],rs[k2]); 64 } 65 void dfs(int k){ 66 if (pos[k])update(r[k],1,n,pos[k]); 67 for(int i=head[k];i!=-1;i=edge[i].nex){ 68 dfs(edge[i].to); 69 merge(r[k],r[edge[i].to]); 70 } 71 } 72 int main(){ 73 scanf("%s%d",s,&m); 74 len[0]=-1; 75 n=strlen(s); 76 x=last=New(); 77 for(int i=0;s[i];i++)add(s[i]-'a',i+1); 78 memset(head,-1,sizeof(head)); 79 for(int i=2;i<=V;i++)add_edge(fa[i],i); 80 dfs(1); 81 x=New(); 82 for(int i=1;i<=m;i++){ 83 scanf("%s%d%d",t,&a,&b); 84 V=x-1; 85 last=New(); 86 for(int j=0,k=1,l=0;t[j];pp[j++]=++l){ 87 int c=t[j]-'a'; 88 add(c,j); 89 while (!query(r[ch[k][c]],1,n,a+l,b)){ 90 if (--l<0)break; 91 if (l==len[fa[k]])k=fa[k]; 92 } 93 if (l<0)k=1; 94 else k=ch[k][c]; 95 } 96 ans=0; 97 for(int j=x;j<=V;j++)ans+=max(0,len[j]-max(len[fa[j]],pp[pos[j]])); 98 printf("%lld\n",ans); 99 } 100 }