Keywords Search(AC自动机模板)

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 68211    Accepted Submission(s): 23017

Problem Description

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

Input

First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.

Output

Print how many keywords are contained in the description.

Sample Input

1

5

she

he

say

shr

her

yasherhs

Sample Output

3

//题意：多个(<=10000)模式串(len<=50)，一个匹配串(len<10^⁶),求匹配数

//AC自动机模板题

指针

 #include <bits/stdc++.h>
 using namespace std;
 #define MX 1000005

 struct Node{
     Node * nex[];
     Node * fail;
     int sum;
     Node(){
         for (int i=;i<;i++) nex[i]=NULL;
         fail=NULL; sum=;
     }
 }*root;

 int n, cnt;
 char str[MX];

 void Init(){
     root=new(Node);
 }

 void Insert(char *s)
 {
     int len = strlen(s);
     Node * p=root;
     for (int i=;i<len;i++)
     {
         int x = s[i]-'a';
         if (p->nex[x]==NULL) p->nex[x]=new(Node);
         p=p->nex[x];
     }
     p->sum++;
 }

 void Build_fail()
 {
     queue<Node *> Q;
     Q.push(root);
     while (!Q.empty())
     {
         Node *p = Q.front(); Q.pop();
         for (int i=;i<;i++)
         {
             if (p->nex[i])
             {
                 if (p==root)
                     p->nex[i]->fail=root;
                 else
                 {
                     Node * tmp = p->fail;
                     while (tmp)
                     {
                         if (tmp->nex[i])
                         {
                             p->nex[i]->fail=tmp->nex[i];
                             break;
                         }
                         tmp = tmp->fail;
                     }
                     if (tmp==NULL) p->nex[i]->fail=root; //
                 }
                 Q.push(p->nex[i]);
             }
         }
     }
 }

 void AC_auto(char *s)
 {
     int len = strlen(s);
     Node *p=root, *tmp;
     for (int i=;i<len;i++)
     {
         int x = s[i]-'a';
         while (p->nex[x]==NULL && p!=root) p=p->fail;
         p=p->nex[x];
         if (!p) p=root;
         tmp = p;
         while (tmp)
         {
             if (tmp->sum>=)
             {
                 cnt+=tmp->sum;
                 tmp->sum=-;
             } else break;
             tmp=tmp->fail;
         }
     }
 }

 void _delete(Node *p)
 {
     for (int i=;i<;i++)
     {
         if (p->nex[i])
             _delete(p->nex[i]);
     }
     delete(p);
 }

 int main()
 {
     int T;
     scanf("%d",&T);
     while (T--)
     {
         scanf("%d",&n);
         Init();
         for (int i=;i<=n;i++)
         {
             scanf("%s",str);
             Insert(str);
         }
         scanf("%s",str);
         cnt = ;
         Build_fail();
         AC_auto(str);
         printf("%d\n",cnt);
         _delete(root);
     }
     return ;
 }