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 = 2*1e4+10;

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][26],l[N],ans[N];

void clear() {
    sz=0; last=++sz;
    memset(ch,0,sizeof(ch));
    memset(l,0,sizeof(l));
}
void add(int x) {
    int c=s[x]-'a';
    int p=last,np=++sz; last=np;
    l[np]=l[p]+1;
    for(;p&&!ch[p][c];p=fa[p]) ch[p][c]=np;
    if(!p) fa[np]=1;
    else {
        int q=ch[p][c];
        if(l[p]+1==l[q]) fa[np]=q;
        else {
            int nq=++sz; l[nq]=l[p]+1;
            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=0;
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) x=x*10+c-'0',c=getchar();
    return x;
}
int main() {
    int T,i,j; T=read();
    while(T--) {
        scanf("%s",s); n=read();
        for(i=1;i<=n;i++) {
            que[i].l=read(),que[i].r=read();
            que[i].num=i;
        }
        sort(que+1,que+n+1);
        memset(ans,0,sizeof(ans));
        j=que[1].l;
        for(i=1;i<=n;i++) {
            if(i==1&&que[i].l==que[i-1].l)
                for(j;j<=que[i].r;j++) add(j-1);
            else {
                clear();
                for(j=que[i].l;j<=que[i].r;j++) add(j-1);
            }
            for(j=1;j<=sz;j++)
                ans[que[i].num]+=l[j]-l[fa[j]];
        }
        for(int i=1;i<=n;i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}