HDU2222Keywords Search ac自动机

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本文介绍了一种基于AC自动机的图像检索系统，通过匹配关键词与图像描述，实现高效精准的图像搜索功能。系统首先将关键词构建成AC自动机，然后在图像描述中查找这些关键词，最终返回匹配度最高的图像。

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http://acm.hdu.edu.cn/showproblem.php?pid=2222

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

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
using namespace std;
const int maxnode=500000+100;
const int sigma_size=26;
int ans;
struct AC_Automata
{
    int ch[maxnode][sigma_size];
    int val[maxnode];//用于保存当前单词输入了几遍,当用find扫描文本的时候,只要遇到该单词就+val[i]且让vis[i]=false,使得下次遇到该模板不会重复计算
    int f[maxnode];
    int last[maxnode];
    bool vis[maxnode];//非单词节点vis=0,单词节点vis=1.如果用find找到了单词i节点,那么vis=0.
    int sz;
    void init()
    {
        sz=1;
        memset(ch[0],0,sizeof(ch[0]));
        vis[0]=false;
        last[0]=f[0]=val[0]=0;
    }
    void insert(char *s)
    {
        int n=strlen(s),u=0;
        for(int i=0;i<n;i++)
        {
            int id=s[i]-'a';
            if(ch[u][id]==0)
            {
                vis[sz]=false;
                ch[u][id]=sz;
                memset(ch[sz],0,sizeof(ch[sz]));
                val[sz++]=0;
            }
            u=ch[u][id];
        }
        vis[u]=true;
        val[u]++;
    }
    void print(int i)
    {
        if(val[i])
        {
            if(vis[i])
            {
                ans+=val[i];
                vis[i]=false;
            }
            print(last[i]);
        }
    }
    void find(char *s)
    {
        int n=strlen(s),j=0;
        for(int i=0;i<n;i++)
        {
            int id=s[i]-'a';
            while(j && ch[j][id]==0) j=f[j];
            j=ch[j][id];
            if(val[j]) print(j);
            else if(last[j]) print(last[j]);
        }
    }
    void getFail()
    {
        queue<int> q;
        f[0]=0;
        for(int i=0;i<sigma_size;i++)
        {
            int u=ch[0][i];
            if(u)
            {
                last[u]=f[u]=0;
                q.push(u);
            }
        }
        while(!q.empty())
        {
            int r=q.front();q.pop();
            for(int i=0;i<sigma_size;i++)
            {
                int u=ch[r][i];
                if(!u) continue;
                q.push(u);
                int v=f[r];
                while(v && ch[v][i]==0) v=f[v];
                f[u]=ch[v][i];
                last[u] = val[f[u]]?f[u]:last[f[u]];
            }
        }
    }
};
AC_Automata ac;
const int MAXN=1000000+100;
char text[MAXN],word[100];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ans=0;
        ac.init();
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",word);
            ac.insert(word);
        }
        ac.getFail();
        scanf("%s",text);
        ac.find(text);
        printf("%d\n",ans);
    }
    return 0;
}