HDU2222 Keywords Search【AC自动机】

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本文介绍了一种使用AC自动机实现的高效关键词搜索算法，该算法能够在大量文本中快速匹配多个关键词。通过构建AC自动机，算法可以一次性处理所有关键词，避免了单独匹配每个关键词的效率低下问题。文章提供了完整的AC自动机构建过程及搜索流程，并通过一个具体的编程实例展示了其应用。

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 78205 Accepted Submission(s): 27175

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

Author

Wiskey

题目链接：HDU2222 Keywords Search

题解：AC自动机模板题

AC的C++代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<stdlib.h>

using namespace std;
const int N=10000005;
int cnt;

char key[70];
char pattern[N];

struct Node{
	Node *next[26];
	Node *fail;
	int sum; 
};

Node *root;
void insert(char *s)
{
	Node *rt=root;
	for(int i=0;s[i];i++){
		int x=s[i]-'a';
		if(!rt->next[x]){
			Node *node=(struct Node*)malloc(sizeof(struct Node));
			for(int j=0;j<26;j++)
			  node->next[j]=NULL;
			node->fail=NULL;
			node->sum=0;
			rt->next[x]=node;
		}
		rt=rt->next[x];
	}
	rt->sum++;
}

void fail_pointer()
{
	queue<Node*>q;
	q.push(root);
	Node *p,*temp;
	while(!q.empty()){
		temp=q.front();
		q.pop();
		for(int i=0;i<26;i++){
			if(temp->next[i]){
				if(temp==root)
				  temp->next[i]->fail=root;
				else{
					p=temp->fail;
					while(p){
						if(p->next[i]){
							temp->next[i]->fail=p->next[i];
							break;
						}
						p=p->fail;
					}
					if(!p)
					  temp->next[i]->fail=root; 
				}
				q.push(temp->next[i]);
			}
		}
	}
}

void ac(char *s)
{
	Node *rt=root;
	for(int i=0;s[i];i++){
		int x=s[i]-'a';
		while(!rt->next[x]&&rt!=root)
		  rt=rt->fail;
		rt=rt->next[x];
		if(!rt)
		  rt=root;
		Node *temp=rt;
		while(temp!=root){
			if(temp->sum>=0){
				cnt+=temp->sum;
				temp->sum=-1;
			}
			else
			  break;
			temp=temp->fail;
		}
	}
}

int main()
{
	int T,n;
	scanf("%d",&T);
	while(T--){
		root=(struct Node*)malloc(sizeof(struct Node));
		for(int i=0;i<26;i++)
		  root->next[i]=NULL;
		root->fail=NULL;
		root->sum=0;
		scanf("%d",&n);
		getchar();
		for(int i=1;i<=n;i++){
			gets(key);
			insert(key);
		}
		gets(pattern);
		cnt=0;
		fail_pointer();
		ac(pattern);
		printf("%d\n",cnt);
	}
	return 0;
}