hdu4641-K-string(后缀自动机)

Problem 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 S_i,S_i+1... S_j 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 2

abc

2

1 a

2

1 a

2

 

Sample Output

0

1

1

 

题意: 开始时给出一个字符串,给出两种操作,一种是在字符串后面添加一个字符,
另一个是查询出现过K次的字串个数。
解析: 建立后缀自动机,添加一个字符插入即可,对于查询,前面计算过的没必要再算,
直接从当前开始往前面找,已经达到K次的就不管,说明前面已经计算过,现在达到
K次的加进答案。

 

代码

#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int maxn=500005;
int N,M,K;
struct SAM
{
    int ch[maxn][26];
    int pre[maxn],step[maxn];
    int last,id,ans;
    int num[maxn];
    void init()
    {
        ans=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;

        while(np!=-1&&num[np]<K) //没有达到K次的就加1
        {
            num[np]++;
            if(num[np]>=K) ans+=step[np]-step[pre[np]]; //加上答案
            np=pre[np];
        }
    }
}sam;
char S[50005];
int main()
{
    while(scanf("%d%d%d",&N,&M,&K)!=EOF)
    {
        scanf("%s",S);
        sam.init();
        for(int i=0;i<N;i++) sam.Insert(S[i]-'a');
        int type;
        char c;
        while(M--)
        {
            scanf("%d",&type);
            if(type==1){ scanf(" %c",&c); sam.Insert(c-'a'); }
            else printf("%d\n",sam.ans);
        }
    }
    return 0;
}