HDU 2222 AC自动机模板

最新推荐文章于 2024-04-26 20:01:35 发布

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本文介绍了一种使用AC自动机实现的高效关键词搜索算法。该算法能够在大量文本中快速定位并统计出指定关键词出现的次数，适用于搜索引擎、图片描述匹配等场景。文章通过一个具体的代码示例详细展示了AC自动机的构建过程及关键词匹配流程。

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 60573 Accepted Submission(s): 19961

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

AC自动机模板,大概就是构造一个类似KMP的next数组的fail指针.然后每次匹配失败时候直接转跳fail指针所指的位置.(之前的模板有点问题,已经修正)

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<queue>
using namespace std;
const int maxn = 5e5+100;
int val[maxn],go[maxn][26],fail[maxn],rec[maxn],sum[maxn];
struct ac_auto
{
  int root,cnt;
  int create()
  {
    cnt++;
    memset(go[cnt],0,sizeof(go[cnt]));
    val[cnt]=0;
    return cnt;
  }
  void init()
  {
    cnt=0;
    root=create();
  }
  void Insert(char s[])
  {
    int len=strlen(s);
    int now=root;
    for (int i=0;i<len;i++)
    {
       int ch=s[i]-'a';
       if (!go[now][ch]) go[now][ch]=create();
       now=go[now][ch];
    }
    val[now]+=1;
  }
  void build()
  {
    static queue<int>q;
    while (!q.empty()) q.pop();
    q.push(root);
    while (!q.empty())
    {
      int u=q.front(); q.pop();
      for (int i=0;i<26;i++)
      {
        if (go[u][i]){
           int y=go[u][i],v=fail[u];
           for (;v&&go[v][i]==0;v=fail[v]);
           fail[y]=v?go[v][i]:root;
           sum[y]=val[y]+sum[fail[y]];
           q.push(y);
        }
      }
    }
  }
  int get(int x)
  {
    int res=0;
    while (x)
    {
      res+=val[x];
      val[x]=0;
      x=fail[x];
    }
    return res;
  }
  int match(char s[])
  {
    int point=1,res=0;
    int now=root,len=strlen(s);
    for (int i=0;i<len;i++)
    {
      int ch=s[i]-'a';
      for (;now&&!go[now][ch];now=fail[now]);
      now=now?go[now][ch]:root;
      if(val[now]) rec[now]=1;
      res+=get(now);
    }
    return res;
  }
}ac;
char st[1000005];
int main()
{

  int T,N;
  scanf("%d",&N);
  while (N--){
    scanf("%d",&T);
    ac.init();
    while (T--)
    {
      char str[100];
      scanf("%s",str);
      ac.Insert(str);
   }
    ac.build();
    scanf("%s",st);
    printf("%d\n",ac.match(st));
  }
  //printf("%d\n",ac.bfs());
}