HDOJ 2222 Keywords Search

ac自动机模板题

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25993    Accepted Submission(s): 8464

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

 

Author

Wiskey

 

Recommend

lcy

 

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

const int kind=26;

struct node

{

    node* fail;

    node* next[kind];

    int cont;

    node()

    {

        fail=NULL;

        cont=0;

        memset(next,0,sizeof(next));

    }

}*q[500001];

char keyworld[51];

char str[1000001];

int head,tail;

void insert(char*str,node* root)

{

    node *p=root;

    int i=0,index;

    while(str)

    {

        index=str-'a';

        if(p->next[index]==NULL) p->next[index]=new node();

        p=p->next[index];

        i++;

    }

    p->cont++;

}

void build_ac(node* root)

{

    root->fail=NULL;

    q[head++]=root;

    while(head!=tail)

    {

        node* temp=q[tail++];

        node* p=NULL;

        for(int i=0;i<26;i++)

        {

            if(temp->next!=NULL)

            {

                if(temp==root)

                    temp->next->fail=root;

                else

                {

                    p=temp->fail;

                    while(p!=NULL)

                    {

                        if(p->next!=NULL)

                        {

                            temp->next->fail=p->next;

                            break;

                        }

                        p=p->fail;

                    }

                    if(p==NULL)

                        temp->next->fail=root;

                }

                q[head++]=temp->next;

            }

        }

    }

}

int query(node* root)

{

    int i=0,cnt=0,index;

    node* p=root;

    while(str)

    {

        index=str-'a';

        while(p->next[index]==NULL&&p!=root)  p=p->fail;

        p=p->next[index];

        p=(p==NULL)?root:p;

        node* temp=p;

        while(temp!=root&&temp->cont!=-1)

        {

            cnt+=temp->cont;

            temp->cont=-1;

            temp=temp->fail;

        }

        i++;

    }

    return cnt;

}

int main()

{

    int n,T;

    scanf("%d",&T);

while(T--)

{

    head=tail=0;

    node* root=new node();

    scanf("%d",&n);

    getchar();

    while(n--)

    {

        gets(keyworld);

        insert(keyworld,root);

    }

    build_ac(root);

    scanf("%s",str);

    printf("%d\n",query(root));

}

    return 0;

}