hdu2222之AC自动机入门

 

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25799    Accepted Submission(s): 8421

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

 

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<cstring>

#include<string>

#include<queue>

#include<algorithm>

#include<map>

#include<iomanip>

#define INF 9999999

using namespace std;



const int MAX=1000000+10;

char a[60],b[MAX];



struct TrieNode{

	int num;//记录单词的数量

	TrieNode *next[26],*fail;//下一个节点和失败指针

	TrieNode(){

		num=0;

		fail=0;

		memset(next,0,sizeof next);

	} 

}*root;



void InsertNode(char *a){

	int k=0;

	TrieNode *p=root;

	while(a[k]){

		if(!p->next[a[k]-'a'])p->next[a[k]-'a']=new TrieNode;

		p=p->next[a[k++]-'a'];

	}

	++p->num;

}



void Build_AC(){

	int k=0;

	TrieNode *p=root,*next;

	queue<TrieNode *>q;

	q.push(p);

	while(!q.empty()){//一层一层的构造fail 

		p=q.front();

		q.pop();

		for(int i=0;i<26;++i){

			if(p->next[i]){

				next=p->fail;

				while(next){

					if(next->next[i]){p->next[i]->fail=next->next[i];break;}

					next=next->fail;

				}

				if(!next)p->next[i]->fail=root;

				q.push(p->next[i]);

			}

		}

	}

}



int SearchTrie(char *a){

	int k=0,sum=0;

	TrieNode *p=root,*next; 

	while(a[k]){

		while(!p->next[a[k]-'a'] && p != root)p=p->fail;

		p=p->next[a[k++]-'a'];

		if(!p)p=root;

		next=p;

		while(next != root && next->num != -1){//不能用0来判断是否已经寻找过 

			sum+=next->num;

			next->num=-1;//这里注意一定要赋值为负数,不能为0 

			next=next->fail;

		}

	}

	return sum;

}



void Free(TrieNode *p){

	for(int i=0;i<26;++i)if(p->next[i])Free(p->next[i]);

	delete p;

}



int main(){

	int T,n;

	cin>>T;

	while(T--){

		root=new TrieNode;

		cin>>n;

		for(int i=0;i<n;++i){

			cin>>a;

			InsertNode(a);//插入单词 

		}

		Build_AC();//构造失败指针,也就是建立AC自动机的精髓

		cin>>b;

		cout<<SearchTrie(b)<<endl;//查询含有的单词数量

		Free(root);

	}

	return 0;

}

/* 

10

2

abcdef

bcd

abcdef

1

h

hhhhh

5

bhea

her

he

h

ha

bhera

5

bhea

her

he

h

ha

bhera

*/