SDNU 1116.Keywords Search AC自动机裸题

最新推荐文章于 2020-07-03 23:23:35 发布

陈年风褛

最新推荐文章于 2020-07-03 23:23:35 发布

阅读量591

点赞数

分类专栏： SDNU

本文链接：https://blog.youkuaiyun.com/lmhacm/article/details/78480594

版权

SDNU 专栏收录该内容

46 篇文章

订阅专栏

本文介绍了一种使用AC自动机解决关键词搜索问题的方法。通过构建字典树并结合KMP算法，高效地实现了在大量文本中搜索多个关键词的功能。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1116.Keywords Search

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 an integer 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
5
she
he
say
shr
her

yasherhs

Sample Output

一道AC自动机的裸题，题意是说输入一个n，然后输入n个字符串，最后再输入一个字符串，问这个字符串包含前面的几个字符串，这类问题的解决就是用AC自动机来解决。AC自动机其实就是先构建一棵字典树，然后用KMP算法来查找匹配，具体的就不细说了，给个链接http://blog.youkuaiyun.com/liu940204/article/details/51347064

下面AC代码：（因为不想用指针所以用的kuangbin板子，没用前面链接里那个板子...）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
struct Trie
{
    int next[500005][26],fail[500005],en[500005];
    int root,L;
    int newnode()
    {
        int i;
        for(i=0;i<26;i++)
        {
            next[L][i]=-1;
        }
        en[L++]=0;
        return L-1;
    }
    void init()
    {
        L=0;
        root=newnode();
    }
    void inser(char buf[])
    {
        int i;
        int len=strlen(buf);
        int now=root;
        for(i=0;i<len;i++)
        {
            if(next[now][buf[i]-'a']==-1)
                next[now][buf[i]-'a']=newnode();
            now=next[now][buf[i]-'a'];
        }
        en[now]++;
    }
    void build()
    {
        int i;
        queue<int>Q;
        fail[root]=root;
        for(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 i;
            int now=Q.front();
            Q.pop();
            for(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);
        int now=root;
        int res=0;
        int i;
        for(i=0;i<len;i++)
        {
            now=next[now][buf[i]-'a'];
            int temp=now;
            while(temp!=root)
            {
                res+=en[temp];
                en[temp]=0;
                temp=fail[temp];
            }
        }
        return res;
    }
    /*void debug()
    {
        int i,j;
        for(i=0;i<L;i++)
        {
            cout<<"id = "<<i<<" fail = "<<fail[i]<<"en = "<<en[i]<<endl;
            cout<<"chi = ";
            for(j=0;j<26;j++)
            {
                cout<<next[i][j];
            }
            cout<<endl;
        }
    }*/
};

char s[1000005];
Trie ac;

int main()
{
    int n;
    int i;
    while(scanf("%d",&n)!=EOF)
    {
        ac.init();
        for(i=0;i<n;i++)
        {
            scanf("%s",s);
            ac.inser(s);
        }
        ac.build();
        scanf("%s",s);
        cout<<ac.query(s)<<endl;
    }
    return 0;
}