HDU 4641 K-string （SAM）

Description

Given a string S. K-string is the sub-string of S and it appear in the S at least K times.It means there are at least K different pairs (i,j) so that Si,Si+1… Sj equal to this K-string. Given m operator or query:1.add a letter to the end of S; 2.query how many different K-string currently.For each query ,count the number of different K-string currently.

 

Input

The input consists of multiple test cases.

Each test case begins with a line containing three integers n, m and K(1<=n,K<=50000,1<=m<=200000), denoting the length of string S, the number of operator or question and the least number of occurrences of K-string in the S.

The second line consists string S,which only contains lowercase letters.

The next m lines describe the operator or query.The description of the operator looks as two space-separated integers t c (t = 1; c is lowercase letter).The description of the query looks as one integer t (t = 2).

 

Output

For each query print an integer — the number of different K-string currently.

 

Sample Input

3 5 2abc21 a21 a2

 

Sample Output

011

 

题意

对于一个长度为 n 的字符串有两种操作：

1. 向字符串的末尾增加一个字符 c
2. 查询串中至少出现 m 次的子串有多少个

 

思路

我们用原串建立后缀自动机，显然第一种操作即一次 insert 。

对于第二种操作，我们考虑每一次的 insert ，对新加入节点 u−>S 的状态路径 num[np]+1 ，若此时的 num[np]>=m 则说明当前状态下所有子串的出现次数都已达到 m ，因此 ans+=step[np]−step[pre[np]] 。

最终的 ans 即为结果。

 

AC 代码

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn = 5e5+10;LL ans, K;struct SAM{    int ch[maxn][26];    int pre[maxn], step[maxn];    int last, id;    int num[maxn];    void init()    {        last = id = 0;        memset(ch[0], -1, sizeof(ch[0]));        pre[0] = -1;        step[0] = 0;    }    void insert(int c)    {        int p = last, np = ++id;        step[np] = step[p] + 1;        memset(ch[np], -1, sizeof(ch[np]));        num[np] = 0;        while (p != -1 && ch[p][c] == -1)            ch[p][c] = np, p = pre[p];        if (p == -1)            pre[np] = 0;        else        {            int q = ch[p][c];            if (step[q] != step[p] + 1)            {                int nq = ++id;                memcpy(ch[nq], ch[q], sizeof(ch[q]));                num[nq] = num[q];                step[nq] = step[p] + 1;                pre[nq] = pre[q];                pre[np] = pre[q] = nq;                while (p != -1 && ch[p][c] == q)                    ch[p][c] = nq, p = pre[p];            }            else                pre[np] = q;        }        last = np;        /* diy */        while (np != -1 && num[np] < K)        {            num[np]++;            if (num[np] >= K)                ans += step[np] - step[pre[np]];            np = pre[np];        }        /* end diy */    }} sam;char str[maxn];int main(){    int n, q;    while (~scanf("%d%d%I64d%*c", &n, &q, &K))    {        sam.init();        ans = 0;        gets(str);        for (int i = 0; i < n; i++)            sam.insert(str[i] - 'a');        for (int i = 0; i < q; i++)        {            int op;            scanf("%d%*c", &op);            if (op == 1)            {                char c;                c = getchar();                sam.insert(c - 'a');            }            else                printf("%I64d\n", ans);        }    }    return 0;}