给定输入文本和k个单词数组arr []，查找输入文本中所有单词的所有出现。令n为文本的长度， m为所有单词的总数字符，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，那么我们需要一个一个地搜索text []中的所有单词。这使我们的总时间复杂度为O(n +长度(word [0])+ O(n +长度(word [1])+ O(n +长度(word [2]])+…O(n +长度( word [k-1])。时间复杂度可以写成O(n * k + m) 。
Aho-Corasick算法查找O(n + m + z)时间中的所有单词，其中z是文本中单词出现的总数。 Aho–Corasick字符串匹配算法构成了原始Unix命令fgrep的基础。
- 处理：在arr []中建立所有单词的自动机。自动机主要具有以下三个功能：
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.
- 匹配：遍历内置自动机上的给定文本以查找所有匹配的单词。
预处理：
- 我们首先为所有单词构建一个Trie(或关键字树)。
特里
- 这部分填充goto g [] []中的条目，并输出o []。
- 接下来，我们将Trie扩展为自动机以支持线性时间匹配。
- 此部分填充失败f []中的条目并输出o []。
去 ：
我们建造特里。对于所有在根部都没有边缘的字符，我们在根部添加一条边缘。
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。
下面是Aho-Corasick算法的实现
C++
// C++ program for implementation of Aho Corasick algorithm
// for string matching
using namespace std;
#include
// 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 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;
} Java
// Java program for implementation of
// Aho Corasick algorithm for String
// matching
import java.util.*;
class GFG{
// Max number of states in the matching
// machine. Should be equal to the sum
// of the length of all keywords.
static int MAXS = 500;
// Maximum number of characters
// in input alphabet
static 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.
static int []out = new int[MAXS];
// FAILURE FUNCTION IS IMPLEMENTED USING f[]
static int []f = new int[MAXS];
// GOTO FUNCTION (OR TRIE) IS
// IMPLEMENTED USING g[][]
static int [][]g = new int[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.
static int buildMatchingMachine(String arr[], int k)
{
// Initialize all values in output function as 0.
Arrays.fill(out, 0);
// Initialize all values in goto function as -1.
for(int i = 0; i < MAXS; i++)
Arrays.fill(g[i], -1);
// Initially, we just have the 0 state
int states = 1;
// Convalues for goto function, i.e., fill g[][]
// This is same as building a Trie for arr[]
for(int i = 0; i < k; ++i)
{
String word = arr[i];
int currentState = 0;
// Insert all characters of current
// word in arr[]
for(int j = 0; j < word.length(); ++j)
{
int ch = word.charAt(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
Arrays.fill(f, -1);
// Failure function is computed in
// breadth first order
// using a queue
Queue q = new LinkedList<>();
// 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.add(g[0][ch]);
}
}
// Now queue has states 1 and 3
while (!q.isEmpty())
{
// Remove the front state from queue
int state = q.peek();
q.remove();
// 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.add(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.
static 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.
static 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.length(); ++i)
{
currentState = findNextState(currentState,
text.charAt(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)) > 0)
{
System.out.print("Word " + arr[j] +
" appears from " +
(i - arr[j].length() + 1) +
" to " + i + "\n");
}
}
}
}
// Driver code
public static void main(String[] args)
{
String arr[] = { "he", "she", "hers", "his" };
String text = "ahishers";
int k = arr.length;
searchWords(arr, k, text);
}
}
// This code is contributed by Princi Singh C#
// C# program for implementation of
// Aho Corasick algorithm for String
// matching
using System;
using System.Collections.Generic;
class GFG{
// Max number of states in the matching
// machine. Should be equal to the sum
// of the length of all keywords.
static int MAXS = 500;
// Maximum number of characters
// in input alphabet
static 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.
static int[] outt = new int[MAXS];
// FAILURE FUNCTION IS IMPLEMENTED USING f[]
static int[] f = new int[MAXS];
// GOTO FUNCTION (OR TRIE) IS
// IMPLEMENTED USING g[,]
static int[,] g = new int[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.
static int buildMatchingMachine(String[] arr, int k)
{
// Initialize all values in output function as 0.
for(int i = 0; i < outt.Length; i++)
outt[i] = 0;
// Initialize all values in goto function as -1.
for(int i = 0; i < MAXS; i++)
for(int j = 0; j < MAXC; j++)
g[i, j] = -1;
// Initially, we just have the 0 state
int states = 1;
// Convalues for goto function, i.e., fill g[,]
// This is same as building a Trie for []arr
for(int i = 0; i < k; ++i)
{
String word = arr[i];
int currentState = 0;
// Insert all characters of current
// word in []arr
for(int j = 0; j < word.Length; ++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
outt[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
for(int i = 0; i < MAXC; i++)
f[i] = 0;
// Failure function is computed in
// breadth first order
// using a queue
Queue q = new Queue();
// 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.Enqueue(g[0, ch]);
}
}
// Now queue has states 1 and 3
while (q.Count != 0)
{
// Remove the front state from queue
int state = q.Peek();
q.Dequeue();
// 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
outt[g[state, ch]] |= outt[failure];
// Insert the next level node
// (of Trie) in Queue
q.Enqueue(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.
static 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.
static 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.Length; ++i)
{
currentState = findNextState(currentState,
text[i]);
// If match not found, move to next state
if (outt[currentState] == 0)
continue;
// Match found, print all matching
// words of []arr
// using output function.
for(int j = 0; j < k; ++j)
{
if ((outt[currentState] & (1 << j)) > 0)
{
Console.Write("Word " + arr[j] +
" appears from " +
(i - arr[j].Length + 1) +
" to " + i + "\n");
}
}
}
}
// Driver code
public static void Main(String[] args)
{
String[] arr = { "he", "she", "hers", "his" };
String text = "ahishers";
int k = arr.Length;
searchWords(arr, k, text);
}
}
// This code is contributed by Amit Katiyar 输出：
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