hdu2222（AC自动机）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2222

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 87731    Accepted Submission(s): 30561

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

 题意：给出n个关键字字符串，然后给出一个字符串s，问有多少个关键字字符串出现在字符串s。

看到这道题，发现这是一道字符串匹配的题，那么很容易想到KMP来解，然而当关键字足够多的时候，那么肯定超时，用trie树也一样，这时我们需要一个新的算法来解决这个问题了：AC自动机。

这里给出一篇关于AC自动机的博客吧：https://bestsort.cn/2019/04/28/402/

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+5;
int tre[maxn][30];
int sum[maxn];
char ch1[maxn];
char ch[105];
queue<int> que;
int fail[maxn];
void insert(char * ch,int & cnt){
	int rt=0;
	int len=strlen(ch);
	for(int i=0;i<len;i++){
		int tem=ch[i]-'a';
		if(tre[rt][tem]==0){
			tre[rt][tem]=++cnt;	
		}
		rt=tre[rt][tem];
	}
	sum[rt]++;
	return ;
}
void getfail(){
	while(!que.empty())que.pop();
	int now=0;
	for(int i=0;i<26;i++){
		if(tre[now][i]==0)continue;
		que.push(tre[now][i]);
	}
	while(!que.empty()){
		now=que.front();
		que.pop();
//		cout<<now<<endl;
		for(int i=0;i<26;i++){
			if(tre[now][i]!=0){
				fail[tre[now][i]]=tre[fail[now]][i];
				que.push(tre[now][i]);
			}
			else{
				tre[now][i]=tre[fail[now]][i];
			}
		}
	}
//	puts("YES");
}
int query(char * ch){
	int len=strlen(ch);
	int ans=0;
	int now=0;
	for(int i=0;i<len;i++){
		now=tre[now][ch[i]-'a'];
		for(int j=now;j&&sum[j]!=-1;j=fail[j]){
			ans+=sum[j];
			sum[j]=-1;
		}
	}
	return ans;
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n;
		int cnt=0;
		memset(sum,0,sizeof(sum));
		memset(fail,0,sizeof(fail));
		memset(tre,0,sizeof(tre));
		scanf("%d",&n);
		for(int i=0;i<n;i++){
			scanf("%s",ch);
			insert(ch,cnt);
		}
		getfail();
		scanf("%s",ch1);
		printf("%d
",query(ch1));
	}
	return 0;
}