HDU 4622 Reincarnation（SAM）

Problem Description

Now you are back,and have a task to do:
Given you a string s consist of lower-case English letters only,denote f(s) as the number of distinct sub-string of s.
And you have some query,each time you should calculate f(s[l...r]), s[l...r] means the sub-string of s start from l end at r.

 

Input

The first line contains integer T(1<=T<=5), denote the number of the test cases.
For each test cases,the first line contains a string s(1 <= length of s <= 2000).
Denote the length of s by n.
The second line contains an integer Q(1 <= Q <= 10000),denote the number of queries.
Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n), denote a query.

 

Output

For each test cases,for each query,print the answer in one line.

 

Sample Input

2
bbaba
5
3 4
2 2
2 5
2 4
1 4
baaba
5
3 3
3 4
1 4
3 5
5 5

 

Sample Output

3
1
7
5
8
1
3
8
5
1

Hint

I won't do anything against hash because I am nice.Of course this problem has a solution that don't rely on hash.

 

Author

WJMZBMR

 

Source

2013 Multi-University Training Contest 3

 

ps：代码自带大常数=-=

 #include<cstdio>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 using namespace std;

 const int N = *1e4+;

 struct QN {
     int l,r,num;
     bool operator < (const QN& rhs) const {
         return l<rhs.l || (l==rhs.l&&r<rhs.r);
     }
 }que[N];
 char s[N];
 int n,sz,last;
 int fa[N],ch[N][],l[N],ans[N];

 void clear() {
     sz=; last=++sz;
     memset(ch,,sizeof(ch));
     memset(l,,sizeof(l));
 }
 void add(int x) {
     int c=s[x]-'a';
     int p=last,np=++sz; last=np;
     l[np]=l[p]+;
     for(;p&&!ch[p][c];p=fa[p]) ch[p][c]=np;
     if(!p) fa[np]=;
     else {
         int q=ch[p][c];
         if(l[p]+==l[q]) fa[np]=q;
         else {
             int nq=++sz; l[nq]=l[p]+;
             memcpy(ch[nq],ch[q],sizeof(ch[q]));
             fa[nq]=fa[q];
             fa[np]=fa[q]=nq;
             for(;q==ch[p][c];p=fa[p]) ch[p][c]=nq;
         }
     }
 }
 int read() {
     char c=getchar(); int x=;
     while(!isdigit(c)) c=getchar();
     while(isdigit(c)) x=x*+c-'',c=getchar();
     return x;
 }
 int main() {
     int T,i,j; T=read();
     while(T--) {
         scanf("%s",s); n=read();
         for(i=;i<=n;i++) {
             que[i].l=read(),que[i].r=read();
             que[i].num=i;
         }
         sort(que+,que+n+);
         memset(ans,,sizeof(ans));
         j=que[].l;
         for(i=;i<=n;i++) {
             if(i==&&que[i].l==que[i-].l)
                 for(j;j<=que[i].r;j++) add(j-);
             else {
                 clear();
                 for(j=que[i].l;j<=que[i].r;j++) add(j-);
             }
             for(j=;j<=sz;j++)
                 ans[que[i].num]+=l[j]-l[fa[j]];
         }
         for(int i=;i<=n;i++)
             printf("%d\n",ans[i]);
     }
     return ;
 }