HDU 4622 Reincarnation （查詢一段字符串的不同子串個數，后綴自動機）

Reincarnation

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 843    Accepted Submission(s): 283

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

 

 

 

 

剛剛學了下后綴自動機，然后把這題先A掉。

 

這題就是查詢一個區間內的不同子串的個數。

如果單單是求一個字符串的不同子串個數，無論是后綴數組還是后綴自動機都非常容易實現。

 

N<=2000.

我是用后綴自動機預處理出所有區間的不同子串個數。

建立n次后綴自動機。

 

后綴自動機要理解其含義，從起點到每個點的不同路徑，就是不同的子串。

到每一個點，不同路徑，其實就是以這個點為最后一個字符的后綴，長度是介於(p->fa->len,p->len]之間的，個數也就清楚了。

而且這個其實是動態變化的，每加入一個字符，就可以知道新加了幾個不同子串。

加個pos，記錄位置，這樣就很容易預處理了。

 

 

學了一天的后綴自動，在世界冠軍cxlove的博客中一直看SAM，終於有點理解了，Orz,太神奇了。

 

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

const int CHAR = 26;
const int MAXN = 2020;
struct SAM_Node
{
    SAM_Node *fa,*next[CHAR];
    int len;
    int id,pos;
    SAM_Node(){}
    SAM_Node(int _len)
    {
        fa = 0;
        len = _len;
        memset(next,0,sizeof(next));
    }
};
SAM_Node SAM_node[MAXN*2], *SAM_root, *SAM_last;
int SAM_size;
SAM_Node *newSAM_Node(int len)
{
    SAM_node[SAM_size] = SAM_Node(len);
    SAM_node[SAM_size].id = SAM_size;
    return &SAM_node[SAM_size++];
}
SAM_Node *newSAM_Node(SAM_Node *p)
{
    SAM_node[SAM_size] = *p;
    SAM_node[SAM_size].id = SAM_size;
    return &SAM_node[SAM_size++];
}
void SAM_init()
{
    SAM_size = 0;
    SAM_root = SAM_last = newSAM_Node(0);
    SAM_node[0].pos = 0;
}
void SAM_add(int x,int len)
{
    SAM_Node *p = SAM_last, *np = newSAM_Node(p->len+1);
    np->pos = len;
    SAM_last = np;
    for(;p && !p->next[x];p = p->fa)
        p->next[x] = np;
    if(!p)
    {
        np->fa = SAM_root;
        return;
    }
    SAM_Node *q = p->next[x];
    if(q->len == p->len + 1)
    {
        np->fa = q;
        return;
    }
    SAM_Node *nq = newSAM_Node(q);
    nq->len = p->len + 1;
    q->fa = nq;
    np->fa = nq;
    for(;p && p->next[x] == q;p = p->fa)
        p->next[x] = nq;
}
void SAM_build(char *s)
{
    SAM_init();
    int len = strlen(s);
    for(int i = 0;i < len;i++)
        SAM_add(s[i] - 'a',i+1);
}

int Q[MAXN][MAXN];
char str[MAXN];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",str);
        int n = strlen(str);
        memset(Q,0,sizeof(Q));
        for(int i = 0;i < n;i++)
        {
            SAM_init();
            for(int j = i;j < n;j++)
            {
                SAM_add(str[j]-'a',j-i+1);
            }
            for(int j = 1;j < SAM_size;j++)
            {
                Q[i][SAM_node[j].pos-1+i]+=SAM_node[j].len - SAM_node[j].fa->len;
            }
            for(int j = i+1;j < n;j++)
                Q[i][j] += Q[i][j-1];
        }
        int M;
        int u,v;
        scanf("%d",&M);
        while(M--)
        {
            scanf("%d%d",&u,&v);
            u--;v--;
            printf("%d\n",Q[u][v]);
        }
    }
    return 0;
}