Hdu2222——Keywords Search（AC自动机模板题）

最新推荐文章于 2024-07-28 10:45:00 发布

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本文通过一个具体的案例，深入浅出地介绍了AC自动机在关键词搜索中的应用。文章首先概述了问题背景，随后详细展示了AC自动机的实现过程，包括初始化、插入关键字、预处理以及查找等关键步骤。

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Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 76055 Accepted Submission(s): 26231

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

题解

没什么好说的，AC自动机裸题。

代码

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<queue>
using namespace std;
queue<int> q;
const int N=1e6+5;
int bo[N],ch[N][30],p[N],tot,ans;
char st[N];
void init()
{
    memset(bo,0,sizeof(bo)); tot=1; ans=0;
    for (int i=0; i<26; i++) ch[0][i]=1,ch[1][i]=0;
}
void insert(char *st)
{
    int n=strlen(st),u=1;
    for (int i=0; i<n; i++){
        int c=st[i]-'a';
        if (c<0) c+=32;
        if (!ch[u][c]) ch[u][c]=++tot,memset(ch[tot],0,sizeof(ch[tot]));
        u=ch[u][c];
    }
    bo[u]++;
}
void pre()
{
    while (!q.empty()) q.pop();
    q.push(1); p[1]=0;
    while (!q.empty()){
        int u=q.front(); q.pop();
        for (int i=0; i<26; i++){
            if (!ch[u][i]) ch[u][i]=ch[p[u]][i];
            else {
                q.push(ch[u][i]);
                p[ch[u][i]]=ch[p[u]][i];
            }
        }
    }
}
void find(char *st)
{
    int u=1,n=strlen(st),k;
    for (int i=0; i<=n; i++){
        int c=st[i]-'a';
        k=ch[u][c];
        while (k>1){
            ans+=bo[k];
            bo[k]=0; k=p[k];
        }
        u=ch[u][c];
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while (T--){
        init(); int n;
        scanf("%d",&n);
        for (int i=1; i<=n; i++) scanf("%s",st),insert(st);
        scanf("%s",st);
        pre();
        find(st);
        printf("%d\n",ans);
    }
    return 0;
}

miao~~~