Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18800    Accepted Submission(s): 6269

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       这题就是AC自动机。 AC自动机参考：
http://www.cnblogs.com/destinydesigner/archive/2009/10/15/1584191.html
http://www.cppblog.com/mythit/archive/2009/04/21/80633.html       总结了下，AC自动机并不难哈~~~

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int MAX=26;
typedef struct Trie_Node
{
    struct Trie_Node *fail;//失败指针
    struct Trie_Node *next[MAX];//26子节点
    int count;//单词最后一个结点的计数
}Trie;
Trie *Q[500010];//队列，用于bfs构造
char keyword[55];
char str[1000010];
int head,tail;//队列的头和尾

void insert(Trie *root,char *word)
{
    Trie *p=root;
    int i=0;
    while(word[i]!='\0')
    {
        if(p->next[word[i]-'a']==NULL)
        {
            Trie *temp=new Trie;
            for(int j=0;j<MAX;j++)
                temp->next[j]=NULL;
            temp->count=0;
            temp->fail=NULL;
            p->next[word[i]-'a']=temp;
        }
        p=p->next[word[i]-'a'];
        i++;
    }
    p->count++;
}
void build_ac(Trie *root)
{
    root->fail=NULL;
    head=tail=0;
    Q[head++]=root;
    while(head!=tail)
    {
        Trie *temp=Q[tail++];
        Trie *p=NULL;
        for(int i=0;i<MAX;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(Trie *root)
{
    int i=0;
    int cnt=0;
    int len=strlen(str);
    Trie *p=root;
    int index;
    while(str[i])
    {
        index=str[i]-'a';
        while(p->next[index]==NULL&&p!=root)p=p->fail;
        p=p->next[index];
        if(p==NULL)p=root;
        Trie *temp=p;
        while(temp!=root)
        {
            cnt+=temp->count;
            temp->count=0;
            temp=temp->fail;
        }
        i++;
    }
    return cnt;
}
void del(Trie *root)
{
    for(int i=0;i<MAX;i++)
      if(root->next[i]!=NULL)
        del(root->next[i]);
    free(root);
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int T;
    int n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        Trie *root=new Trie;
        root->count=0;
        root->fail=NULL;
        for(int i=0;i<MAX;i++)
              root->next[i]=NULL;
        for(int i=0;i<n;i++)
        {
            scanf("%s",&keyword);
            insert(root,keyword);
        }
        build_ac(root);
        scanf("%s",&str);
        printf("%d\n",query(root));
        del(root);//时间卡得紧就不删除空间
    }
    return 0;
}