AC自动机(AC automation)

　　字典树+KMP

　　参考自: http://www.cppblog.com/mythit/archive/2009/04/21/80633.html

　　

 const int MAXN = ; //字典大小

 //定义结点
 struct node{
     node* fail;
     node* child[MAXN];
     int count;
     node(){
         fail = NULL;
         count = ;
         memset(child, NULL, sizeof(child));
     }
 };

 //将词插入字典树
 void insert(node* root, char* str){
     node* p = root;
     int i = ;
     while(str[i]){
         int index = str[i] - 'a';
         if(p->child[index] == NULL) p->child[index] = new node();
         p = p->child[index];
         ++i;
     }
     p->count++;
 }

 //BFS构造失配指针/构建ACautomation
 //由当前结点构造子结点失配指针
 void build_acautomation(node* root){
     queue<node*> q;
     root->fail = NULL;
     q.push(root);
     while(!q.empty()){
         node* temp = q.front();
         node* p = NULL;
         for(int i = ; i < MAXN; ++i){
             if(temp->child[i] != NULL){
                 if(temp == root) temp->child[i]->fail = root;
                 else{
                     p = temp->fail;
                     while(p != NULL){
                         if(p->child[i] != NULL){
                             temp->child[i]->fail = p->child[i];
                             break;
                         }
                         p = p->fail;
                     }
                     if(p == NULL) temp->child[i]->fail = root;
                 }
                 q.push(temp->child[i]);
             }
         }
     }
 }

 //ACautomation匹配
 //计算当前进行匹配的字符串及其后缀子串的个数
 //count因题而异
 int query(node* root){
     int i = , cnt = , index;
     node* p = root;
     while(str[i]){
         index = str[i]-'a';
         while(p->child[index] == NULL && p != root) p = p->fail;
         p = p->child[index];
         p = (p == NULL ? root : p);
         node* temp = p;
         while(temp != root && temp->count != -){
             cnt += temp->count;
             temp->count = -;
             temp = temp->fail;
         }
         ++i;
     }
     return cnt;
 }

　　继续学习..........准备做题(VJ kuangbin AC自动机专题)