A KUANGBIN~~AC automatic version of the sub, with annotations, the role of counting.

The dictionary tree of the cock silk ... Drunk ...
#include<cstdio>#include<algorithm>#include<iostream>#include<cstring>#include<queue>#include<cmath>#include<cstdlib>UsingNamespace Std;struct Trie{int Next[500010][26],fail[500010],isend[500010];int root, L;IntNewNode(){For(int I=0; I<26; I++)This string of jumps empties next[L][i]=-1; Isend[L++]=0;Not a node layerReturn L-1;Returns the ordinal of this point}voidInit(){L=0; Root=NewNode();Give the root a point}voidInsert(Char buf[]){int Len=Strlen(buf);int now= root;For(int I=0; I< Len; I++){If(Next[Now][buf[I]-A]==-1)This character is not present in the current dictionary layer next[Now][buf[I]-A]=NewNode();Point to generate a new point, and this has the character now= Next[Now][buf[I]-A];Go to the new point to continue operation} isend[Now]++;This layer is a junction layer, labeled}voidBuild(){Queue<Int>q; Fail[Root]= root;The root node is still the root nodeFor(int I=0; I<26; I++)Traversal of the first characterIf(Next[Root][i]==-1)No this character starts with next[Root][i]= root;Jump to RootElseWith the beginning of this character{Fail[Next[Root][i]]= root;The failure pointer for this row bit is root Q.Push(Next[Root][i]);Line into queue} while (! Q.empty ())//the character {int now = Q.front ();//Layer out the first q.pop (); for (int i = 0;i < 26;i++)//For this line if (next[now][i] = = 1)//if the next Line does not have this character next[now][i] = next[fail[now]][i];//His next position, is back to this point the parent node of the next (back) else//If there is this character {fail[next[now][i]]=next[ fail[now]][i];//his next failure pointer to this position is the next Q.push (Next[now][i]) of this point parent node;//The Next line continues}}} int query (char buf[]) {int len = strlen (bu f); int now = root; int res = 0; for (int i = 0;i < len;i++) {now = next[now][buf[i]-' a '], int temp = now, while (temp! = root) {res + = isend[temp]; I Send[temp] = 0; temp = fail[temp]; }} return res; } void Debug () {for (int i = 0;i < l;i++) {printf ("id =%3d,fail =%3d,end =%3d,chi = [", I,fail[i],isend[i]); for (int j = 0;j < 26;j++) printf ("%2d", Next[i][j]); printf ("]\n"); }}};char buf[1000010]; Trie Ac;int Main () {int T; int n; scanf ("%d", &t), while (t--) {scanf ("%d", &n), Ac.init (); for (int i = 0;i < n ; i++) {scanf ("%s", buf), AC. Insert (BUF); } ac.build (); scanf ("%s", buf); printf ("%d\n", Ac.query (BUf)); } return 0;}

Pitch Kuangbin~~ac, with annotations, the role of counting.