HDU 4787 GRE Words Revenge

Description

　　Now Coach Pang is preparing for the Graduate Record Examinations as George did in 2011. At each day, Coach Pang can:
　　 "+w": learn a word w
　　 "?p": read a paragraph p, and count the number of learnt words. Formally speaking, count the number of substrings of p which is a learnt words.
　　Given the records of N days, help Coach Pang to find the count. For convenience, the characters occured in the words and paragraphs are only '0' and '1'.
　
　

Solution

这题网上大部分题解都是错的，只能说数据太水
首先这题用到AC自动机，如果暴力做每一组询问需要getfail一次，但是这样做也能AC.
考虑合并，我们建立两个AC自动机，保证其中一个大小(<sqrt{n}) 每一次对小的进行getfail，复杂度只有(O(sqrt{n}))，如果小的 (size) 达到了 (sqrt{n})，考虑暴力合并，复杂度 (O(sqrt{n}))
注意一个细节，一个单词只能学习一次，所以每加一次需要判断之前是否存在，网上大部分都没有做这个

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <queue>
#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=100005,M=600005;
int m,kase=0,ans=0,B=100,root=0;char s[M*10];
queue<int>q;
struct Ac{
   int size,ch[M][2],fail[M],val[M];
   int newnode(){
      size++;
      ch[size][0]=ch[size][1]=0;
      fail[size]=0;val[size]=0;
      return size;
   }
   void clear(){
      size=-1;newnode();
   }
   void ins(char *S){
      int x,len=strlen(s),p=root;
      for(int i=1;i<len;i++){
         x=S[i]-'0';
         if(ch[p][x])p=ch[p][x];
         else ch[p][x]=newnode(),p=ch[p][x];
      }
      val[p]|=1;
   }
   void getfail(){
      while(!q.empty())q.pop();
      q.push(root);int x,u,v;
      while(!q.empty()){
         x=q.front();q.pop();
         for(int i=0;i<=1;i++){
            if(!ch[x][i])continue;
            u=fail[x];
            while(u && !ch[u][i])u=fail[u];
            if(ch[u][i] && ch[u][i]!=ch[x][i])
               fail[ch[x][i]]=ch[u][i];
            v=ch[x][i];q.push(v);
         }
      }
   }
   int query(char *S){
      int x,len=strlen(S),p=root,ret=0,u;
      for(int i=1;i<len;i++){
         x=S[i]-'0';
         while(p && !ch[p][x])p=fail[p];
         p=ch[p][x];u=p;
         while(u)ret+=val[u],u=fail[u];
      }
      return ret;
   }
   bool check(char *S){
      int len=strlen(S),x,p=root;
      for(int i=1;i<len;i++){
         x=S[i]-'0';
         if(!ch[p][x])return false;
         p=ch[p][x];
      }
      return val[p];
   }
}A,C;
void dfs(int art,int crt){
   int x;
   for(int i=0;i<=1;i++)
      if(C.ch[crt][i]){
         if(!A.ch[art][i])A.ch[art][i]=A.newnode();
         x=A.ch[art][i];A.val[x]|=C.val[C.ch[crt][i]];
         dfs(x,C.ch[crt][i]);
      }
}
char b[N];
void Moveit(){
   int k=ans,len=strlen(s),p=1;
   k%=(len-1);
   for(int i=1;i<=k;i++)b[i]=s[i];
   for(int i=1;k<len-1;i++)s[i]=s[++k];
   for(int i=len-(ans%(len-1));i<len;i++)s[i]=b[p++];
}
void work()
{
   printf("Case #%d:
",++kase);
   scanf("%d",&m);
   A.clear();C.clear();ans=0;
   while(m--){
      scanf("%s",s);
      Moveit();
      if(s[0]=='+'){
         if(A.check(s) || C.check(s))continue;
         C.ins(s);
         if(C.size>B){dfs(0,0);A.getfail();C.clear();}
         else C.getfail();
      }
      else ans=A.query(s)+C.query(s),printf("%d
",ans);
   }
}

int main()
{
	int T;cin>>T;
   while(T--)work();
	return 0;
}