HDU 4622 Reincarnation SAM

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Reincarnation

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 1439    Accepted Submission(s): 503

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.
 

 

Source

2013 Multi-University Training Contest 3

 

Recommend

zhuyuanchen520

 

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后缀自动机啊，真是不明觉厉

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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <climits>
#include <numeric>
#include <vector>
#define sz(x) int(x.size())
using namespace std;

typedef vector<int> VI;
const int maxn = 5000 + 10;

class SuffixAutomaton{
private:
    struct Node{
        Node *suf, *go[26];
        int val;
        Node(){
            suf=NULL;
            val=0;
            memset(go,0,sizeof(go));
        }
        void clear(){
            suf=NULL;
            val=0;
            memset(go,0,sizeof(go));
        }
        int calc(){
            if (suf==0) return 0;
            return val-suf->val;
        }
    };
    Node *root,*last;
    Node nodePool[maxn*2],*cur;
    Node* newNode(){
        Node* res=cur++;
        res->clear();
        return res;
    }
    int tot;
    void extend(int w){
        Node *p=last;
        Node *np=newNode();
        np->val=p->val+1;
        while (p&&!p->go[w]){
            p->go[w]=np;
            p=p->suf;
        }
        if (!p){
            np->suf=root;
            tot+=np->calc();
        }
        else{
            Node *q=p->go[w];
            if (p->val+1==q->val){
                np->suf=q;
                tot+=np->calc();
            }
            else{
                Node *nq=newNode();
                memcpy(nq->go,q->go,sizeof(q->go));
                tot-=p->calc()+q->calc();
                nq->val=p->val+1;
                nq->suf=q->suf;
                q->suf=nq;
                np->suf=nq;
                tot+=p->calc()+q->calc()+np->calc()+nq->calc();
                while (p&&p->go[w]==q){
                    p->go[w]=nq;
                    p=p->suf;
                }
            }
        }
        last = np;
    }
public:
    void init(){
        cur=nodePool;
        root=newNode();
        last=root;
    }
    VI getSubString(char s[]){
        VI v;
        tot=0;
        int len=strlen(s);
        for (int i=0;i<len;i++){
            extend(s[i]-'a');
            v.push_back(tot);
        }
        return v;
    }
}atm;

int ans[maxn][maxn];
char s[maxn];

int main() {
	int T;
	scanf("%d",&T);
	while (T--){
	    scanf("%s",s);
	    int len=strlen(s);
	    for (int i=0;i<len;i++){
            atm.init();
            VI v=atm.getSubString(s+i);
            for (int j=0;j<sz(v);j++) ans[i][i+j]=v[j];
	    }
	    int nQ;
        scanf("%d",&nQ);
        while (nQ--){
            int l,r;
            scanf("%d%d",&l,&r);
            printf("%d
",ans[l-1][r-1]);
        }
	}
	return 0;
}

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