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  • AC自动机(AC automation)

      字典树+KMP

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

      

     1 const int MAXN = 26; //字典大小
     2 
     3 //定义结点
     4 struct node{
     5     node* fail;
     6     node* child[MAXN];
     7     int count;
     8     node(){
     9         fail = NULL;
    10         count = 0;
    11         memset(child, NULL, sizeof(child));
    12     }
    13 };
    14 
    15 //将词插入字典树
    16 void insert(node* root, char* str){
    17     node* p = root;
    18     int i = 0;
    19     while(str[i]){
    20         int index = str[i] - 'a';
    21         if(p->child[index] == NULL) p->child[index] = new node();
    22         p = p->child[index];
    23         ++i;
    24     }
    25     p->count++;
    26 }
    27 
    28 //BFS构造失配指针/构建ACautomation
    29 //由当前结点构造子结点失配指针
    30 void build_acautomation(node* root){
    31     queue<node*> q;
    32     root->fail = NULL;
    33     q.push(root);
    34     while(!q.empty()){
    35         node* temp = q.front();
    36         node* p = NULL;
    37         for(int i = 0; i < MAXN; ++i){
    38             if(temp->child[i] != NULL){
    39                 if(temp == root) temp->child[i]->fail = root;
    40                 else{
    41                     p = temp->fail;
    42                     while(p != NULL){
    43                         if(p->child[i] != NULL){
    44                             temp->child[i]->fail = p->child[i];
    45                             break;
    46                         }
    47                         p = p->fail;
    48                     }
    49                     if(p == NULL) temp->child[i]->fail = root;
    50                 }
    51                 q.push(temp->child[i]);
    52             }
    53         }
    54     }
    55 }
    56 
    57 //ACautomation匹配
    58 //计算当前进行匹配的字符串及其后缀子串的个数
    59 //count因题而异
    60 int query(node* root){
    61     int i = 0, cnt = 0, index;
    62     node* p = root;
    63     while(str[i]){
    64         index = str[i]-'a';
    65         while(p->child[index] == NULL && p != root) p = p->fail;
    66         p = p->child[index];
    67         p = (p == NULL ? root : p);
    68         node* temp = p;
    69         while(temp != root && temp->count != -1){
    70             cnt += temp->count;
    71             temp->count = -1;
    72             temp = temp->fail;
    73         }
    74         ++i;
    75     }
    76     return cnt;
    77 }

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

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  • 原文地址:https://www.cnblogs.com/book-book/p/5357669.html
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