HDU 2222 Keywords Search (AC自动机)

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

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

 

/*
 * HDU 2222 Keywords Search
 * AC自动机模板题
 * 
 * 动态内存版
 */

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
using namespace std;

const int MAXN = 100000+10;
const int MAX = 26;
typedef struct Trie_Node
{
    struct Trie_Node *fail;//失配指针
    struct Trie_Node *next[MAX];//子节点
    int cnt;
    Trie_Node(){fail=NULL;cnt=0;memset(next,NULL,sizeof(next))};
}Trie;
Trie *Q[500010];//队列，用于bfs构造
int front,tail;//队列头和尾

void Trie_Init(Trie* root)
{
    for(int i=0;i<MAX;i++) root->next[i]=NULL;
    root->cnt=0;
    root->fail=NULL;
}
void Trie_Insert(Trie* root,char* str)
{
    Trie* p=root;
    int len=strlen(str);
    for(int i=0;i<len;i++)
    {
        int id=str[i]-'a';
        if(p->next[id]==NULL)
        {
            Trie* t=new Trie;
            Trie_Init(t);
            p->next[id]=t;
        }
        p=p->next[id];
    }
    p->cnt++;
}
void Build_AC(Trie* root)
{
    front=tail=0;
    Q[front++]=root;
    while(front!=tail)
    {
        Trie* t=Q[tail++];
        Trie* p=NULL;
        for(int i=0;i<MAX;i++)
        {
            if(t->next[i]==NULL) continue;
            if(t==root) t->next[i]->fail=root;
            else
            {
                p=t->fail;
                while(p!=NULL)
                {
                    if(p->next[i]!=NULL)
                    {
                        t->next[i]->fail=p->next[i];
                        break;
                    }
                    p=p->fail;
                }
                if(p==NULL) t->next[i]->fail=root;
            }
            Q[front++]=t->next[i];
        }
    }
}
int Trie_Query(Trie* root,char* str)
{
    int res=0;
    Trie* p=root;
    int len=strlen(str);
    for(int i=0;i<len;i++)
    {
        int id=str[i]-'a';
        while(p->next[id]==NULL&&p!=root) p=p->fail;
        p=p->next[id];
        if(p==NULL) p=root;
        Trie* t=p;
        while(t!=root)
        {
            res+=t->cnt;
            t->cnt=0;
            t=t->fail;
        }
    }
    return res;
}
void Trie_Del(Trie* root)
{
    for(int i=0;i<MAX;i++)
        if(root->next[i]!=NULL)
            Trie_Del(root->next[i]);
    free(root);
}
char str[MAXN];

int main()
{
    int T;
    int n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        Trie* root=new Trie;
        Trie_Init(root);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",str);
            Trie_Insert(root,str);
        }
        Build_AC(root);
        scanf("%s",str);
        printf("%d\n",Trie_Query(root,str));
        Trie_Del(root);//时间紧就不删除空间
    }
}