hdu2222Keywords Search AC自动机模板题

DescriptionIn 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. InputFirst 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. OutputPrint how many keywords are contained in the description. Sample Input

15shehesayshrheryasherhs 

Sample Output

3

Sample Output

3

#include<iostream>#include<cstdio>#include<cstring>#include<string>using namespace std;#define N 500010char str[1000010],keyword[51];int head,tail;struct node{    node *fail;    node *next[26];    int count;    node()    {        fail=NULL;        count=0;        for(int i=0;i<26;i++) next[i]=NULL;    }}*q;node *root;void insert(char *str){    int temp,len;    node *p=root;    len=strlen(str);    for(int i=0;i<len;i++)    {        temp=str[i]-'a';        if(p->next[temp]==NULL) p->next[temp]=new node();        p=p->next[temp];    }    p->count++;}void build_ac(){    q[tail++]=root;    while(head!=tail)    {        node *p=q[head++];        node *temp=NULL;        for(int i=0;i<26;i++)        {            if(p->next[i]!=NULL)            {                if(p==root) p->next[i]->fail=root;                else                {                    temp=p->fail;                    while(temp!=NULL)                    {                        if(temp->next[i]!=NULL)                        {                            p->next[i]->fail=temp->next[i];                            break;                        }                        temp=temp->fail;                    }                    if(temp==NULL) p->next[i]->fail=root;                }                q[tail++]=p->next[i];            }        }    }}int query(){    int index,len,result;    node *p=root;    result=0;    len=strlen(str);    for(int i=0;i<len;i++)    {        index=str[i]-'a';        while(p->next[index]==NULL&&p!=root) p=p->fail;        p=p->next[index];        if(p==NULL) p=root;        node *temp=p;        while(temp!=root&&temp->count!=-1)        {            result+=temp->count;            temp->count=-1;            temp=temp->fail;        }    }    return result;}int main(){  //  freopen("cin.txt","r",stdin);    int ncase,num;    scanf("%d",&ncase);    while(ncase--)    {        head=tail=0;        root=new node();        scanf("%d",&num);        getchar();        for(int i=0;i<num;i++)        {            gets(keyword);            insert(keyword);        }        build_ac();        scanf("%s",str);        printf("%d\n",query());    }    return 0;}