AC自动机模板——HDU2222 Keywords Search

最新推荐文章于 2020-03-09 18:52:11 发布

原创最新推荐文章于 2020-03-09 18:52:11 发布 · 365 阅读

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#acm #程序竞赛 #字符串 #AC自动机

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本文介绍了一种利用AC自动机进行高效关键词匹配的方法。通过构建自动机来处理多个关键词的快速检索，适用于大规模文本数据的关键词匹配场景，如搜索引擎等。

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Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 62125 Accepted Submission(s): 20552

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

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思路：建立字典树升级为自动机查找就行这道题是AC自动机入门题可以当作模板

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;
struct ACa
{
    int next[500050][26];
    int fail[500050],end[500050];
    int root,cnt;//根结点，计数器
    int newnode()
    {
        for(int i=0;i<26;i++)
        {
            next[cnt][i]=-1;//子结点设为空
        }
        end[cnt++]=0;//以该结点结束的串数
        return cnt-1;
    }
    int clear(){//初始化
        cnt=0;//结点数重置
        root=newnode();//初始化根结点 //完成后root=0
    }
    int insert(char c[])
    {
        int now=root;
        int len=strlen(c);
        for(int i=0;i<len;i++)
        {
            if(next[now][c[i]-'a']==-1)//如果对应结点未建立，添加结点
                next[now][c[i]-'a']=newnode();
            now=next[now][c[i]-'a'];//否则沿结点继续
        }
        end[now]++;
    }
    void build()
    {
        queue<int>q;
        fail[root]=root;
        for(int i=0;i<26;i++){
            if(next[root][i]==-1)
                next[root][i]=root;
            else{//有结点则入队
                fail[next[root][i]]=root;
                q.push(next[root][i]);
            }
        }
        while(!q.empty())
        {
            int now=q.front();
            q.pop();
            for(int i=0;i<26;i++){
                if(next[now][i]==-1)
                  next[now][i]=next[fail[now]][i];//链接到fail指针对应结点的后面
                else{
                  fail[next[now][i]]=next[fail[now]][i];
                  q.push(next[now][i]);
                }
            }
        }
    }
    int query(char c[])
    {//查询
        int len=strlen(c);
        int now=root;
        int res=0;
        for(int i=0;i<len;i++)
            {
            now=next[now][c[i]-'a'];
            int temp=now;
            while(temp!=root)
            {
                res+=end[temp];
                end[temp]=0;
                temp=fail[temp];
            }
        }
        return res;
    }
};
char buf[1000500];
ACa ac;
int main(){
    int T,n;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        ac.clear();
        int i;
        for(i=1;i<=n;i++){
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        scanf("%s",buf);
        printf("%d\n",ac.query(buf));
    }
    return 0;
}