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 ;

 }

巴特西

HDU 4622 Reincarnation（SAM）

最新文章

热门文章