HDU2222 Keywords Search [AC自动机]

Posted 2020-12-29 cytus

　　题目传送门

Keywords Search

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

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　　分析：

　　水一波$AC$自动机裸题。

　　直接上$AC$自动机模板就行了，记得清空数组，其他别想太多就好。

　　Code：

 

//It is made by HolseLee on 14th Aug 2018
//HDU2222
#include<bits/stdc++.h>
using namespace std;

const int N=5e5+7;
int T,n,m,ans,tot,fail[N],team[N],t[N][26],val[N];
char s[N*2];

void ins(char *ch)
{
    int len=strlen(ch),root=0,id;
    for(int i=0;i<len;++i){
        id=ch[i]-‘a‘;
        if(!t[root][id])
        t[root][id]=++tot;
        root=t[root][id];
    }
    val[root]++;
}

void getfail()
{
    int root=0,head=0,tail=0;;
    for(int i=0;i<26;++i){
        if(t[root][i]){
            fail[t[root][i]]=0;
            team[++tail]=t[root][i];
        }
    }
    while(head<tail){
        root=team[++head];
        for(int i=0;i<26;++i){
            if(t[root][i]){
                fail[t[root][i]]=t[fail[root]][i];
                team[++tail]=t[root][i];
            }
            else t[root][i]=t[fail[root]][i];
        }
    }
}

void quary()
{
    int root=0,id;
    for(int i=0;i<m;++i){
        id=s[i]-‘a‘;
        root=t[root][id];
        for(int j=root;j&&~val[j];j=fail[j])
        ans+=val[j],val[j]=-1;
    }
}

int main()
{
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        ans=tot=0;
        memset(t,0,sizeof(t));
        memset(val,0,sizeof(val));
        memset(fail,0,sizeof(fail));
        memset(team,0,sizeof(team));
        for(int i=1;i<=n;++i){
            scanf("%s",s);ins(s);
        }
        getfail();
        scanf("%s",s);
        m=strlen(s);
        quary();
        printf("%d
",ans);
    }
    return 0;
}