字符串（后缀自动机）：HDU 4622 Reincarnation

Reincarnation

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 3194    Accepted Submission(s): 1184

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.

 

　　程立杰出的题目？

　　sam有个性质就是对于新加一个字符，产生的新子串个数为len[lst]-len[fa[lst]]。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int maxn=;
 int fa[maxn],ch[maxn][];
 int len[maxn],lst,cnt;
 int dp[maxn][maxn];
 char s[maxn];
 struct Opt{
     void Init(){
         memset(ch,,sizeof(ch));
         lst=cnt=;
     }

     int Insert(int c){
         int p=lst,np=lst=++cnt;len[np]=len[p]+;
         while(p&&ch[p][c]==)ch[p][c]=np,p=fa[p];
         if(!p)fa[np]=;
         else{
             int q=ch[p][c];
             if(len[q]==len[p]+)fa[np]=q;
             else{
                 int nq=++cnt;len[nq]=len[p]+;
                 memcpy(ch[nq],ch[q],sizeof(ch[q]));
                 fa[nq]=fa[q];fa[q]=fa[np]=nq;
                 while(ch[p][c]==q)ch[p][c]=nq,p=fa[p];
             }
         }
         return len[lst]-len[fa[lst]];
     }
 }SAM;
 int main(){
     int T,Q;
     scanf("%d",&T);
     while(T--){
         scanf("%s",s+);
         int len=strlen(s+);
         for(int i=;i<=len;i++){
             SAM.Init();
             for(int j=i;j<=len;j++)
                 dp[i][j]=dp[i][j-]+SAM.Insert(s[j]-'a');
         }
         scanf("%d",&Q);
         while(Q--){
             int l,r;
             scanf("%d%d",&l,&r);
             printf("%d\n",dp[l][r]);
         }
     }
     return ;
 }