hdu 2222 Keywords Search

最新推荐文章于 2025-08-15 23:10:13 发布

转载最新推荐文章于 2025-08-15 23:10:13 发布 · 70 阅读

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原文链接：http://www.cnblogs.com/L-Memory/p/7039783.html

文章标签：

#java

本文详细介绍了使用AAC自动机进行关键词搜索的方法，并通过一个具体的图片检索系统的实例来展示如何实现关键词匹配。文章提供了完整的代码实现，包括关键词插入、失败指针构造及匹配过程。

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Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 63090 Accepted Submission(s): 20917

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

Recommend

lcy

/*
AAC自动机模板 寻找字串在模式串中出现个数
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>

#define N 500007

using namespace std;

queue<int>q;
int n,t,sz=1,len;
int v[N],fail[N],trie[N][27];
char s[51],a[N*20]; 
bool mark[N];

void insert()
{
    int root=1;
    for(int i=0;i<len;i++)
    {
        int id=s[i]-'a'+1;
        if(!trie[root][id]) trie[root][id]=++sz;
        root=trie[root][id];
    }
    v[root]++;
}

void buildfail()
{
    for(int i=1;i<=26;i++) trie[0][i]=1;
    q.push(1);
    while(!q.empty())
    {
        int now=q.front();
        for(int i=1;i<=26;i++)
        {
            if(!trie[now][i]) continue;
            q.push(trie[now][i]);int x=fail[now];
            while(!trie[x][i]) x=fail[x];
            fail[trie[now][i]]=trie[x][i];
        }
        q.pop();
    }
}

void work()
{
    int root=1,ans=0;
    for(int i=0;i<len;i++)
    {
        int id=a[i]-'a'+1;mark[root]=true;
        while(!trie[root][id]) root=fail[root];
        root=trie[root][id];
        if(!mark[root])//每个点都找一遍 
        {
            int x=root;
            while(x)
            {
                ans+=v[x];v[x]=0;x=fail[x];//因为当前后缀等于比它深度浅的前缀，后缀有前缀一定有 
            }
        }
    }printf("%d\n",ans);
}

void init()
{
    memset(v,0,sizeof(v));sz=1;
    memset(trie,0,sizeof(trie));
    memset(fail,0,sizeof(fail));
    memset(mark,0,sizeof(mark));
}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        init();scanf("%d",&n);
        for(int i=1;i<=n;i++) 
        {
            scanf("%s",s); len=strlen(s);
            insert();
        }
        buildfail();
        scanf("%s",a);len=strlen(a);
        work();
    }
}