HDU 2222 ----AC自动机

最新推荐文章于 2023-07-09 22:02:16 发布

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最新推荐文章于 2023-07-09 22:02:16 发布

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原文链接：http://www.cnblogs.com/dzzy/p/4907421.html

本文探讨了如何利用AC自动机实现关键词与图像描述的高效匹配。通过输入关键词和图像描述，程序计算并输出匹配的关键词数量。

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

AC自动机，李大神刚刚讲过的内容来一发练习，模板大法好！

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<queue>
using namespace std;

struct Trie
{
    int next[500010][26],fail[500010],end[500010];
    int root,L;
    int newnode()
    {
        for(int i=0;i<26;i++)
        {
            next[L][i]=-1;
        }
        end[L++]=0;
        return L-1;
    }
    void init()
    {
        L=0;
        root=newnode();
    }
    void insert(char buf[])
    {
        int len=strlen(buf);
        int now=root;
        for(int i=0;i<len;i++)
        {
            if(next[now][buf[i]-'a']==-1)
                next[now][buf[i]-'a']=newnode();
            now=next[now][buf[i]-'a'];
        }
        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
            {
                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];
                else
                {
                    fail[next[now][i]]=next[fail[now]][i];
                    q.push(next[now][i]);
                }
            }
        }
    }    
    int query(char buf[])
    {
        int len=strlen(buf),now=root,res=0;
        for(int i=0;i<len;i++)
        {
            now=next[now][buf[i]-'a'];
            int temp=now;
            while(temp!=root)
            {
                res+=end[temp];
                end[temp]=0;
                temp=fail[temp];
            }
        }
        return res;
    }
};

char buf[1000010];
Trie ac;
int main()
{
    //freopen("in.txt","r",stdin);
    int T,n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        ac.init();
        for(int i=0;i<n;i++)
        {
            scanf("%s",buf);
            ac.insert(buf);
        }
        ac.build();
        scanf("%s",buf);
        printf("%d\n",ac.query(buf));
    }
    return 0;
}