hdu----1686 Oulipo (ac自动机)

Oulipo

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6227    Accepted Submission(s):
2513

Problem Description

The French author Georges Perec (1936–1982) once wrote
a book, La disparition, without the letter 'e'. He was a member of the Oulipo
group. A quote from the book:

Tout avait Pair normal, mais tout
s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain,
l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait
au roman : stir son tapis, assaillant à tout instant son imagination,
l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un
non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la
raison : tout avait l’air normal mais…

Perec would probably have scored
high (or rather, low) in the following contest. People are asked to write a
perhaps even meaningful text on some subject with as few occurrences of a given
“word” as possible. Our task is to provide the jury with a program that counts
these occurrences, in order to obtain a ranking of the competitors. These
competitors often write very long texts with nonsense meaning; a sequence of
500,000 consecutive 'T's is not unusual. And they never use spaces.

So we
want to quickly find out how often a word, i.e., a given string, occurs in a
text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite
strings over that alphabet, a word W and a text T, count the number of
occurrences of W in T. All the consecutive characters of W must exactly match
consecutive characters of T. Occurrences may overlap.

 

Input

The first line of the input file contains a single
number: the number of test cases to follow. Each test case has the following
format:

One line with the word W, a string over {'A', 'B', 'C', …, 'Z'},
with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
One line
with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤
1,000,000.

 

Output

For every test case in the input file, the output
should contain a single number, on a single line: the number of occurrences of
the word W in the text T.

 

Sample Input

3

BAPC

BAPC

AZA

AZAZAZA

VERDI

AVERDXIVYERDIAN

 

Sample Output

1

3

0

 

Source

华东区大学生程序设计邀请赛_热身赛

 

   ac自动机做法：

   

 

代码：

 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 #include<iostream>
 #include<queue>

 using namespace std;
 const int maxn=;

 struct node{

  node  *child[maxn];
  node * fail ;  //失败指针
  int tail ;

 };
 void init(node * tmp){
     for(int i=;i<maxn;i++)
       tmp->child[i]=NULL;
       tmp->fail=NULL;
       tmp->tail=;
 }

 //建立一颗字典树

 void Buildtree(const char ss [] ,node * root){
      node *cur;
    for(int i=; ss[i] ; i++){
         int no =ss[i]-'A';
          if(root->child[no]==NULL){
               cur = new node ;
               init(cur);
               root->child[no]=cur;
         }
       root = root->child[no];
     }
     root->tail++;
 }

 //构造失败指针
 void BuildFail(node * root){

   queue<node *> tmp;
   tmp.push(root);
   node *cur,*po;
   while(!tmp.empty()){
     cur = tmp.front();
      tmp.pop();       //出队列
      for(int i= ; i<maxn ; i++){
         if( cur->child[i] != NULL ){
            if(cur==root){
               cur->child[i]->fail=root;      //指向树根
             }else{
                  po = cur ;
                while(po->fail){
                  if(po->fail->child[i]){
                     cur->child[i]->fail=po->fail->child[i];
                      break;
                  }
                  po = po->fail;
                }
                if(po->fail==NULL)
                   cur->child[i]->fail=root;
             }
             tmp.push(cur->child[i]);
         }
      }
   }
 }

 //查询
 int Query(char ss [] , node *root){

     node *cur , *tmp;
     cur =root;
     int pos=-,res=;

     for(int i=; ss[i] ;i++){

         pos= ss[i]-'A';
         while(cur->child[pos]==NULL&&cur!=root)
             cur = cur->fail;
         cur = cur->child[pos];
         if(cur==NULL) cur=root;
           tmp = cur;
         while(tmp!=root&&tmp->tail>){
            res+=tmp->tail;
            tmp = tmp->fail;
         }
     }

     return res;

 }
 char sa[],sb[];
 int main(){
  int te;
  node *root;
 scanf("%d",&te);
 while(te--){
    root = new node ;
    init(root);
    scanf("%s %s",sa,sb);
    Buildtree(sa,root);
    BuildFail(root);
    int ans =Query(sb,root);
    printf("%d\n",ans);
 }
 return ;
 }