模式搜索的Aho-Corasick算法如何实现？代码呢？

2021年4月3日19:15:53 发表评论 847 次浏览

给定一个输入文本和一个包含k个单词的数组arr[]，查找输入文本中出现的所有单词。设n为文本长度，m为所有单词中字符总数，即m = length(arr[0]) + length(arr[1]) +…+ length(arr[k-1])。这里k是输入单词的总数。

例子：

Input: text = "ahishers"    
       arr[] = {"he", "she", "hers", "his"}

Output:
   Word his appears from 1 to 3
   Word he appears from 4 to 5
   Word she appears from 3 to 5
   Word hers appears from 4 to 7

如果我们使用像KMP这样的线性时间搜索算法，那么我们需要逐个搜索text[]中的所有单词。这使我们的总时间复杂度为O(n + length(word[0]) + O(n + length(word[1]) + O(n + length(word[2]) +…O(n + length(word[k-1]))。这个时间复杂度可以写成O(n*k + m)。

Aho-Corasick算法在O(n + m + z)时间内查找所有单词，其中z是单词在文本中出现的总次数。Aho-Corasick字符串匹配算法构成了原始Unix命令fgrep的基础。

  1. 处理前：建立一个由arr []中所有单词组成的自动机自动机主要具有三个功能：
Go To :   This function simply follows edges
          of Trie of all words in arr[]. It is
          represented as 2D array g[][] where
          we store next state for current state 
          and character.

Failure : This function stores all edges that are
          followed when current character doesn't
          have edge in Trie.  It is represented as
          1D array f[] where we store next state for
          current state. 

Output :  Stores indexes of all words that end at 
          current state. It is represented as 1D 
          array o[] where we store indexes
          of all matching words as a bitmap for 
          current state.
  1. 匹配：在内置的自动机上遍历给定的文本以找到所有匹配的单词。

预处理：

  1. 我们首先建立一个特里(或关键字树)中的所有单词。
     
模式搜索的Aho-Corasick算法1

Trie

  1. 这部分填充goto g [] []中的条目并输出o []。
  2. 接下来, 我们将Trie扩展为自动机以支持线性时间匹配。
     
模式搜索的Aho-Corasick算法2
  1. 这部分填写失败f[]和输出o[]中的条目。

我们建立单词查找树。对于所有没有根边的字符，我们把根边加回去。

失败：

对于状态s, 我们找到最长的适当后缀, 该后缀是某些模式的适当前缀。这是使用Trie的广度优先遍历完成的。

输出：

对于状态s, 存储所有以s结尾的单词的索引。这些索引存储为按位映射(通过对值进行按位或)。这也使用具有故障的广度优先遍历进行计算。

以下是Aho-Corasick算法的C ++实现

C

// C++ program for implementation of Aho Corasick algorithm
// for string matching
using namespace std;
#include <bits/stdc++.h>
 
// Max number of states in the matching machine.
// Should be equal to the sum of the length of all keywords.
const int MAXS = 500;
 
// Maximum number of characters in input alphabet
const int MAXC = 26;
 
// OUTPUT FUNCTION IS IMPLEMENTED USING out[]
// Bit i in this mask is one if the word with index i
// appears when the machine enters this state.
int out[MAXS];
 
// FAILURE FUNCTION IS IMPLEMENTED USING f[]
int f[MAXS];
 
// GOTO FUNCTION (OR TRIE) IS IMPLEMENTED USING g[][]
int g[MAXS][MAXC];
 
// Builds the string matching machine.
// arr -   array of words. The index of each keyword is important:
//         "out[state] & (1 << i)" is > 0 if we just found word[i]
//         in the text.
// Returns the number of states that the built machine has.
// States are numbered 0 up to the return value - 1, inclusive.
int buildMatchingMachine(string arr[], int k)
{
     // Initialize all values in output function as 0.
     memset (out, 0, sizeof out);
 
     // Initialize all values in goto function as -1.
     memset (g, -1, sizeof g);
 
     // Initially, we just have the 0 state
     int states = 1;
 
     // Construct values for goto function, i.e., fill g[][]
     // This is same as building a Trie for arr[]
     for ( int i = 0; i < k; ++i)
     {
         const string &word = arr[i];
         int currentState = 0;
 
         // Insert all characters of current word in arr[]
         for ( int j = 0; j < word.size(); ++j)
         {
             int ch = word[j] - 'a' ;
 
             // Allocate a new node (create a new state) if a
             // node for ch doesn't exist.
             if (g[currentState][ch] == -1)
                 g[currentState][ch] = states++;
 
             currentState = g[currentState][ch];
         }
 
         // Add current word in output function
         out[currentState] |= (1 << i);
     }
 
     // For all characters which don't have an edge from
     // root (or state 0) in Trie, add a goto edge to state
     // 0 itself
     for ( int ch = 0; ch < MAXC; ++ch)
         if (g[0][ch] == -1)
             g[0][ch] = 0;
 
     // Now, let's build the failure function
 
     // Initialize values in fail function
     memset (f, -1, sizeof f);
 
     // Failure function is computed in breadth first order
     // using a queue
     queue< int > q;
 
      // Iterate over every possible input
     for ( int ch = 0; ch < MAXC; ++ch)
     {
         // All nodes of depth 1 have failure function value
         // as 0. For example, in above diagram we move to 0
         // from states 1 and 3.
         if (g[0][ch] != 0)
         {
             f[g[0][ch]] = 0;
             q.push(g[0][ch]);
         }
     }
 
     // Now queue has states 1 and 3
     while (q.size())
     {
         // Remove the front state from queue
         int state = q.front();
         q.pop();
 
         // For the removed state, find failure function for
         // all those characters for which goto function is
         // not defined.
         for ( int ch = 0; ch <= MAXC; ++ch)
         {
             // If goto function is defined for character 'ch'
             // and 'state'
             if (g[state][ch] != -1)
             {
                 // Find failure state of removed state
                 int failure = f[state];
 
                 // Find the deepest node labeled by proper
                 // suffix of string from root to current
                 // state.
                 while (g[failure][ch] == -1)
                       failure = f[failure];
 
                 failure = g[failure][ch];
                 f[g[state][ch]] = failure;
 
                 // Merge output values
                 out[g[state][ch]] |= out[failure];
 
                 // Insert the next level node (of Trie) in Queue
                 q.push(g[state][ch]);
             }
         }
     }
 
     return states;
}
 
// Returns the next state the machine will transition to using goto
// and failure functions.
// currentState - The current state of the machine. Must be between
//                0 and the number of states - 1, inclusive.
// nextInput - The next character that enters into the machine.
int findNextState( int currentState, char nextInput)
{
     int answer = currentState;
     int ch = nextInput - 'a' ;
 
     // If goto is not defined, use failure function
     while (g[answer][ch] == -1)
         answer = f[answer];
 
     return g[answer][ch];
}
 
// This function finds all occurrences of all array words
// in text.
void searchWords(string arr[], int k, string text)
{
     // Preprocess patterns.
     // Build machine with goto, failure and output functions
     buildMatchingMachine(arr, k);
 
     // Initialize current state
     int currentState = 0;
 
     // Traverse the text through the nuilt machine to find
     // all occurrences of words in arr[]
     for ( int i = 0; i < text.size(); ++i)
     {
         currentState = findNextState(currentState, text[i]);
 
         // If match not found, move to next state
         if (out[currentState] == 0)
              continue ;
 
         // Match found, print all matching words of arr[]
         // using output function.
         for ( int j = 0; j < k; ++j)
         {
             if (out[currentState] & (1 << j))
             {
                 cout << "Word " << arr[j] << " appears from "
                      << i - arr[j].size() + 1 << " to " << i << endl;
             }
         }
     }
}
 
// Driver program to test above
int main()
{
     string arr[] = { "he" , "she" , "hers" , "his" };
     string text = "ahishers" ;
     int k = sizeof (arr)/ sizeof (arr[0]);
 
     searchWords(arr, k, text);
 
     return 0;
}

输出如下：

Word his appears from 1 to 3
Word he appears from 4 to 5
Word she appears from 3 to 5
Word hers appears from 4 to 7

资源：

http://www.cs.uku.fi/~kilpelai/BSA05/lectures/slides04.pdf

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