(字典树+AC自动机) hdu 2222 Keywords Search

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6306    Accepted Submission(s): 2134

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
/*
题目大意：输入信息搜索引擎在一大串图片描述信息中找与输入信息像匹配的的数目
题目解答：字典树+AC自动机

AC自动机 http://blog.youkuaiyun.com/Hashmat/archive/2010/09/23/5902263.aspx
字典树 略

Problem : 2222 ( Keywords Search )     Judge Status : Accepted
RunId : 2987974    Language : C++    Author : wawadimu
Code Render Status : Rendered By HDOJ C++ Code Render Version 0.01 Beta
*/
#include<iostream>
#include<queue>
using namespace std;

#define kind 26
#define maxn 1000010
#define maxlen 51
char res[maxlen];//字典串
char str[maxn];//模式串
struct trie
{
	int cnt;
	trie* fail,*next[kind];
	trie()
	{
		cnt=0;
		fail=NULL;
		memset(next,NULL,sizeof(next));
	}
}*q[500001];//空间开大一点

void insert(char *res,trie *root);//插入单词
void ac_automation(trie* root);//ac自动机
int query(char *str,trie* root);//查找字典序中包含的单词数目

int main()
{
	//freopen("2222.txt","r",stdin);
	int i,j,n;
	int cas;
	scanf("%d",&cas);
	while(cas--)
	{
		trie *root=new trie();
		scanf("%d/n",&n);
		for(i=1;i<=n;i++) 
		{
			scanf("%s/n",res);
			//printf("%s/n",res);
			insert(res,root);
		}		
		scanf("%s/n",str);
		//printf("%s/n",str);
		ac_automation(root);
		int ans=query(str,root);
		printf("%d/n",ans);
	}
	return 0;
}
void insert(char *res,trie *root)
{
	trie *p=root;
	while(*res)
	{
		//cout<<*res<<"  ";
		int index=(*res++)-'a';
		if( p->next[index]==NULL )
			p->next[index]=new trie();
		p=p->next[index];
	}
	p->cnt++;
}
void ac_automation(trie* root)
{
	trie *tmp=NULL,*u=NULL,*p=root;
	p->fail=NULL;
	int front=0,rear=0;
		rear++;
	q[front]=p;
	while(front!=rear)
	{
		u=q[front++];
		for(int i=0;i<kind;i++)
		{
			if(u->next[i]!=NULL)
			{
				if(u==root) u->next[i]->fail=root;
				else
				{
					tmp=u->fail;
					while(tmp!=NULL)
					{
						if(tmp->next[i]!=NULL)
						{
							
							u->next[i]->fail=tmp->next[i];
							tmp=tmp->next[i];
							break;
						}
						tmp=tmp->fail;
					}
					if(tmp==NULL) u->next[i]->fail=root;
				}
				q[rear++]=u->next[i];
			}
		}
	}
}
int query(char *str,trie *root)
{
	trie *tmp,*p=root;
	int cnt=0;
	while(*str)
	{
		int index=(*str++)-'a';
		while(p!=root && p->next[index]==NULL) p=p->fail;
		p=p->next[index];
		if(p==NULL) p=root;
		else
		{
			tmp=p;
			while(tmp!=root && tmp->cnt!=-1)//避免重复增加同样的单词
			{
				cnt+=tmp->cnt;
				tmp->cnt=-1;
				tmp=tmp->fail;
			}
		}
	}
	return cnt;
}