数组版
1 struct Tire{ 2 int next[maxn][27],fail[maxn],dis[maxn]; 3 int root,L; 4 queue<int >q; 5 6 int newnode(){ 7 dis[L] = 0; 8 for(int i = 0; i < 26; i++) 9 next[L][i] = -1; 10 L++; 11 return L-1; 12 } 13 void ini(){ 14 while(q.size()) q.pop(); 15 L = 0,root = newnode(); 16 } 17 int cha(char x){ 18 return x-'a'; 19 } 20 void inser(char buf[]){ 21 int len = strlen(buf); 22 int now = root; 23 for(int i = 0; i < len; i++){ 24 int ta = cha(buf[i]); 25 if(next[now][ta] == -1) 26 next[now][ta] = newnode(); 27 now = next[now][ta]; 28 } 29 dis[now] ++; 30 } 31 void build(){ 32 fail[root] = root; 33 for(int i = 0; i < 26; i++) 34 if(next[root][i] == -1) 35 next[root][i] = root; 36 else{ 37 fail[next[root][i]] = root; 38 q.push(next[root][i]); 39 } 40 while(!q.empty()){ 41 int now = q.front(); 42 q.pop(); 43 for(int i = 0; i < 26; i++){ 44 if(next[now][i] == -1) 45 next[now][i] = next[fail[now]][i]; 46 else{ 47 fail[next[now][i]] = next[fail[now]][i]; 48 q.push(next[now][i]); 49 } 50 } 51 } 52 } 53 int solve(char* str){ 54 int cur = root; 55 int len = strlen(str); 56 int index; 57 int cnt = 0; 58 for(int i = 0; i < len; i++){ 59 index = cha(str[i]); 60 cur = next[cur][index]; 61 int tp = cur; 62 while(tp != root && dis[tp] != -1){ 63 if(dis[tp] > 0){ 64 cnt += dis[tp]; 65 dis[tp] = -1; 66 } 67 tp = fail[tp]; 68 } 69 } 70 return cnt; 71 } 72 };
指针版
1 struct node{ 2 node *fail;//记录匹配失败时要跳到哪个节点 3 node *next[26]; 4 int v;//是否为最后一个节点 5 node(){ 6 v = 0; 7 fail = NULL; 8 memset(next,NULL,sizeof(next)); 9 } 10 }*q[10000*50+5];//队列 11 12 char s1[51];//匹配串 13 char s2[1000005];//母串 14 int head,tail;//队头队尾，注意初始化 15 void build(char *s,node *root){//建一棵字典树 16 node *p = root; 17 int i=0,index; 18 while(s[i]){ 19 index = s[i] - 'a'; 20 if(p->next[index] == NULL) p->next[index] = new node(); 21 p = p->next[index]; 22 i++; 23 } 24 p->v ++;//最后一个节点++，表示出现次数加1 25 } 26 void buildac(node *root){//在字典树上做类似kmp的操作,表示匹配失败时应该跳到什么位置 27 root->fail = NULL; 28 head = tail = 0; 29 q[head++] = root; 30 while(head!=tail){ 31 node *p = q[tail++]; 32 node *temp = NULL; 33 for(int i=0;i<26;i++){ 34 if(p->next[i] != NULL){ 35 q[head++] = p->next[i]; 36 if(p == root) p->next[i]->fail = root; 37 else{ 38 temp = p->fail; 39 while(temp!=NULL){ 40 if(temp->next[i]!=NULL){ 41 p->next[i]->fail = temp->next[i]; 42 break; 43 } 44 temp = temp->fail; 45 } 46 if(temp == NULL) p->next[i]->fail = root; 47 } 48 } 49 } 50 } 51 }