hdu 2222 Keywords Search AC自动机

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26526    Accepted Submission(s): 8651

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

15shehesayshrheryasherhs

 

Sample Output

3

 

Author

Wiskey

 

Recommend

lcy

ac自动机的经典题目，每组数据给定n 个单词 ，然后给定一个字符串yasherhs。问一共有多少单词在这个字符串中出现过。

#include <iostream>#include <cstring>using namespace std;const int kind = 26;struct node{    node *fail;       //失败指针    node *next[kind]; //Tire每个节点的26个子节点（最多26个字母）    int count;        //是否为该单词的最后一个节点    node()            //构造函数初始化    {        fail=NULL;        count=0;        memset(next,NULL,sizeof(next));    }}*q[500001];          //队列，方便用于bfs构造失败指针char keyword[51];     //输入的单词char str[1000001];    //模式串int head,tail;        //队列的头尾指针void insert(char *str,node *root){    node *p=root;    int i=0,index;    while(str[i])    {        index=str[i]-'a';        if(p->next[index]==NULL) p->next[index]=new node();        p=p->next[index];        i++;    }    p->count++;}void build_ac_automation(node *root){    int i;    root->fail=NULL;    q[head++]=root;    while(head!=tail)    {        node *temp=q[tail++];        node *p=NULL;        for(i=0; i<26; i++)        {            if(temp->next[i]!=NULL)            {                if(temp==root) temp->next[i]->fail=root;                else                {                    p=temp->fail;                    while(p!=NULL)                    {                        if(p->next[i]!=NULL)                        {                            temp->next[i]->fail=p->next[i];                            break;                        }                        p=p->fail;                    }                    if(p==NULL) temp->next[i]->fail=root;                }                q[head++]=temp->next[i];            }        }    }}int query(node *root){    int i=0,cnt=0,index,len=strlen(str);    node *p=root;    while(str[i])    {        index=str[i]-'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->count!=-1)        {            cnt+=temp->count;            temp->count=-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(keyword);            insert(keyword,root);        }        build_ac_automation(root);        scanf("%s",str);        printf("%d\n",query(root));    }    return 0;}