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/*===============================* according to the training guide \*===============================*/#include <cstring> #include <queue> #include <cstdio> #include <map> #include <string>using namespace Std;const int Sigma_ SIZE = 26;const int maxnode = 11000;const int maxs = max + 10;map<string,int> ms;struct ahocorasickautomata {int c  H[maxnode][sigma_size];    int F[maxnode];  Fail function int Val[maxnode]; Each end node of a string has a Val int last[maxnode] that is not 0;  The next node of the output list is int cnt[maxs];  int sz;    void init () {sz = 1;    memset (Ch[0], 0, sizeof (ch[0));    memset (CNT, 0, sizeof (CNT));  Ms.clear (); } inline void Clear () {memset (cnt,0,sizeof (CNT));}    Assuming that text is more than just one word, it is often necessary to empty the CNT array//character c number inline int idx (char c) {return C ' a ') every time you find it; Be careful here, assuming you don't have a given range of characters. direct return C;//due to possible negative ... The problem with virus attack is}//insert String.
V must be non 0 void insert (char *s, int v) {int u = 0, n = strlen (s);      for (int i = 0; i < n; i++) {int c = idx (s[i]);        if (!ch[u][c]) {memset (Ch[sz], 0, sizeof (Ch[sz]));        Val[sz] = 0;      CH[U][C] = sz++;     } u = ch[u][c];//u is the position of the first dimension of CH stored in the next node, which is equivalent to the NXT in my trie} val[u] = V;  V is the additional information that best separates each word so that//cnt can record what has appeared and how many times ms[string (s)] = V; }//recursively print all strings ending with node J void print (int j) {if (j) {Cnt[val[j]]++;//val[j] is the number of the word.    Ms stores the corresponding numbers and words, which can be used to print the word "print" (Last[j]);    }}//in T find template int find (char* T) {int n = strlen (t); int j = 0;      Current node number, initial root node for (int i = 0; i < n; i++) {//text string current pointer int c = IDX (T[i]); while (J &&!ch[j][c]) j = f[j];      Walk along the thin edge until you can match j = Ch[j][c]; if (Val[j]) print (j);//To the end of the word else if (Last[j]) print (last[j]);    Got it!    }}//calculate the Fail function void Getfail () {queue<int> q;    F[0] = 0;   Initialize queue for (int c = 0; c < sigma_size; C + +) {int u = ch[0][c];   if (u) {f[u] = 0; Q.push (u); last[u] = 0;} }//because the first character mismatch needs to be matched again,//So the first character points to root (root is the trie entry, no actual meaning)//meaning that all words the first character of the f[] are equal to 0.      The fail pointer to node e points to root to indicate that there is no matching sequence//in BFS sequence to calculate the fail while (!q.empty ()) {int r = Q.front (); Q.pop ();        for (int c = 0; c < sigma_size; C + +) {int u = ch[r][c];        if (!u) continue;        Q.push (U);        int v = f[r]; while (v &&!ch[v][c]) v = f[v];//ch[v][c]==0, it means that there is no continuation of the matching letter edge.        is also unable to continue matching, so continue along the mismatch function walk f[u] = Ch[v][c]; Last[u] = Val[f[u]]?        F[u]: Last[f[u]; LAST[J] Node J goes back along the adapter pointer, and encounters the next word node number//last is to solve after finding a word, see if there are other strings included}}}; Ahocorasickautomata AC;

1, see a range of characters, change sigma_size and IDX function

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