hdu 2222 Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 66013    Accepted Submission(s): 22119

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

 

题意：T组数据，每组数据输入n ，输入n个小字符串 ，最后一行输入一个大字符串，输出它包含多少个小字符串。

AC自动机练习(指针版)

屠龙宝刀点击就送

#include <cstring>
#include <cstdio>
#define N 500010
struct node
{
    int cnt;
    node * next[27],* fail;
}*Q[N],*root;
int T;
node * create()
{
    node * rt=new node;
    rt->cnt=0;
    rt->fail=0;
    memset(rt->next,0,sizeof(rt->next));
    return rt;
}
void ins(char *a)
{
    node * p=root;
    char *q=a;
    while(*q)
    {
        int id=*q-'a'+1;
        if(p->next[id]==NULL) p->next[id]=create();
        p=p->next[id];
        q++;
    }
    p->cnt++;
}
void build()
{
    int head=0,tail=1;
    Q[tail]=root;
    while(head!=tail)
    {
        node * now=Q[++head];
        node * tmp=NULL;
        for(int i=1;i<=26;i++)
        {
            if(now->next[i]!=NULL)
            {
                if(now==root) now->next[i]->fail=root;
                else
                {
                    tmp=now->fail;
                    while(tmp!=NULL)
                    {
                        if(tmp->next[i]!=NULL)
                        {
                            now->next[i]->fail=tmp->next[i];
                            break;
                        }
                        tmp=tmp->fail;
                    }
                    if(tmp==NULL)
                        now->next[i]->fail=root;
                }
                Q[++tail]=now->next[i];
            }
        }
    }
}
int query(char *a)
{
    int ret=0;
    char *q=a;
    node *p=root;
    while(*q)
    {
        int id=*q-'a'+1;
        while(p->next[id]==NULL&&p!=root) p=p->fail;
        p=p->next[id];
        if(p==NULL) p=root;
        node *tmp=p;
        while(tmp!=root&&tmp->cnt!=-1)
        {
            ret+=tmp->cnt;
            tmp->cnt=-1;
            tmp=tmp->fail;
        }
        ++q;
    }
    return ret;
}
int main()
{
    scanf("%d",&T);
    for(int n;T--;)
    {
        root=create();
        char str[55];
        scanf("%d",&n);
        for(;n--;)
        {
            scanf("%s",str);
            ins(str);
        }
        build();
        char key[1000005];
        scanf("%s",key);
        printf("%d
",query(key));
    }
    return 0;
}