AC自动机---Keywords Search

题目网址：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=110773#problem/A

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

 

题意：给了n个模式串，然后给了一个长串，求这个长串的子串中出现了多少个模式串。

 

思路：使用AC自动机的多模式匹配算法，建立trie树，然后使用bfs搜索求所有串中字母的fail指针，然后利用trie树进行模式匹配，求解。

 

代码如下：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 500010
char str[],keyword[];
int head,tail;

struct node
{
    node *fail;
    node *next[];
    int count;
    node()
    {
        fail=NULL;
        count=;
        for(int i=;i<;i++)
        next[i]=NULL;
    }
}*q[N];
node *root;

void insert(char *str) ///建立Trie
{
    int temp,len;
    node *p=root;
    len=strlen(str);
    for(int i=;i<len;i++)
    {
        temp=str[i]-'a';
        if(p->next[temp]==NULL)
           p->next[temp]=new node();
        p=p->next[temp];
    }
    p->count++;
}

void setfail() ///使用bfs初始化fail指针；
{
    q[tail++]=root;
    while(head!=tail)
    {
        node *p=q[head++];
        node *temp=NULL;
        for(int i=;i<;i++)
        if(p->next[i]!=NULL)
        {
            if(p==root) ///首字母的fail必指向根
            p->next[i]->fail=root;
            else
            {
                temp=p->fail; ///失败指针
                while(temp!=NULL) ///2种情况结束：匹配为空or找到匹配
                {
                    if(temp->next[i]!=NULL) ///找到匹配
                    {
                        p->next[i]->fail=temp->next[i];
                        break;
                    }
                    temp=temp->fail;
                }
                if(temp==NULL) ///为空则从头匹配
                    p->next[i]->fail=root;
                }
            q[tail++]=p->next[i]; ///入队
        }
    }
}

int query()
{
    int index,len,result;
    node *p=root;
    result=;
    len=strlen(str);
    for(int i=;i<len;i++)
    {
        index=str[i]- 'a';
        while(p->next[index]==NULL&&p!=root) ///跳转失败指针
        p=p->fail;
        p=p->next[index];
        if(p==NULL)
        p=root;
        node *temp=p; ///p不动，temp计算后缀串
        while(temp!=root&&temp->count!=-)
        {
            result+=temp->count;
            temp->count=-;
            temp=temp->fail;
        }
    }
    return result;
}

int main()
{
    int T, num;
    scanf("%d",&T);
    while(T--)
    {
        head=tail=;
        root = new node();
        scanf("%d", &num);
        getchar();
        for(int i=;i<num;i++)
        {
            gets(keyword);
            insert(keyword);
        }
        setfail();
        scanf("%s",str);
        printf("%d\n",query());
    }
    return ;
}