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  • hdu 2222 Keywords Search

    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
    3
     
    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
    ",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();
        }
    }
    折花枝,恨花枝,准拟花开人共卮,开时人去时。 怕相思,已相思,轮到相思没处辞,眉间露一丝。
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  • 原文地址:https://www.cnblogs.com/L-Memory/p/7039783.html
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