HDU 4641 K-string

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.

solution

还是老套路：利用父节点为最大子集，直接累加儿子出现次数到父亲上面，每一次新加入节点就直接从当前节点跳father，并更新father，如果出现了K次就统计答案，统计方式和上次一样，当前节点所能接受的并且和父亲节点不重合的有 \(len[p]-len[fa[p]]\) 个，注意为了避免重复统计，如果当前节点到了 \(K\)，那么父节点就一定都已经达到了 \(K\)，所以不需要继续跳了

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int N=400005;
int ch[N][27],fa[N],n,m,K,len[N],cnt=1,cur=1,p,size[N];
char s[5],S[N];long long ans=0;
void build(int c,int id){
   p=cur;cur=++cnt;len[cur]=id;
   for(;p && !ch[p][c];p=fa[p])ch[p][c]=cur;
   if(!p)fa[cur]=1;
   else{
      int q=ch[p][c];
      if(len[p]+1==len[q])fa[cur]=q;
      else{
         int nt=++cnt;len[nt]=len[p]+1;
         memcpy(ch[nt],ch[q],sizeof(ch[q]));
         fa[nt]=fa[q];fa[q]=fa[cur]=nt;
         for(;ch[p][c]==q;p=fa[p])ch[p][c]=nt;
         size[nt]=size[q];
      }
   }
}
void upd(){
   p=cur;
   while(p>1){
      if(size[p]>=K)return ;
      size[p]++;
      if(size[p]>=K){ans+=len[p]-len[fa[p]];return ;}
      p=fa[p];
   }
}
void Clear(){
   for(RG int i=0;i<N;i++){
      fa[i]=size[i]=len[i]=0;
      for(RG int j=0;j<=26;j++)ch[i][j]=0;
   }
   cur=1;cnt=1;ans=0;
}
void work()
{
   scanf("%s",S+1);
   for(int i=1;i<=n;i++){
      build(S[i]-'a',i);
      upd();
   }
   int flag;
   for(int i=1;i<=m;i++){
      scanf("%d",&flag);
      if(flag==2)printf("%lld\n",ans);
      else{
         scanf("%s",s);
         build(s[0]-'a',++n);upd();
      }
   }
}

int main()
{
    while(~scanf("%d%d%d",&n,&m,&K))work(),Clear();
	return 0;
}