HDU 2222 Keywords Search AC自动机

Keywords Search

Problem Description

 

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey also wants to bring this feature to his image retrieval system.
Every image have a long description, when users type some keywords to find the image, the system will match the keywords with description of image and show the image which the most keywords be matched.
To simplify the problem, giving you a description of image, and some keywords, you should tell me how many keywords will be match.

 

Input

 

First line will contain one integer means how many cases will follow by.
Each case will contain two integers N means the number of keywords and N keywords follow. (N <= 10000)
Each keyword will only contains characters 'a'-'z', and the length will be not longer than 50.
The last line is the description, and the length will be not longer than 1000000.

 

Output

 

Print how many keywords are contained in the description.

 

Sample Input

 

1 5 she he say shr her yasherhs

 

Sample Output

 

3

第一道AC自动机；

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N=1e6+10,mod=1e9+7,inf=2e9+10;

typedef struct Trie_node{
  struct Trie_node *fail;//失败指针
  struct Trie_node *next[27];//26子节点
  int counts;//单词最后一个结点的计数
  Trie_node(){
    fail = NULL;
    counts = 0;
    memset(next,NULL,sizeof(next));
  }
}Trie;
Trie *q[500010];
char keyword[55];//输入的单词
char str[1000006];//输入的模式串
int head,tail;//队列的头尾指针

//将模式串插入trie树当中
void insert(char *str,Trie_node *root) {
    Trie_node *p = root;
    int i = 0, index;
    while(str[i]) {
        index = str[i] - 'a';
        if(p->next[index] == NULL) p->next[index] = new Trie_node();
        p = p->next[index];
        ++i;
    }
    p->counts++;
}

//建立失配fail
void build_ac_automation(Trie_node *root) {
    int i;
    head = tail = 0;
    root->fail = NULL;
    q[tail++] = root;
    while(head!=tail) {
        Trie_node *temp = q[head++];
        Trie_node *p = NULL;
        for(i = 0; i < 26; ++i) {
            if(temp->next[i]!=NULL) {
                if(temp == root) temp->next[i]->fail = root;
                else {
                    p = temp->fail;
                    while(p!=NULL) {
                        if(p->next[i]!=NULL){
                            temp->next[i]->fail=p->next[i];
                            break;
                        }
                        p=p->fail;
                    }
                    if(p == NULL) temp->next[i]->fail = root;
                }
                q[tail++] = temp->next[i];
            }
        }
    }
}
int query(char *s,Trie_node *root){
    int len = strlen(s),cnt = 0;
    Trie_node *p = root;
    for(int i = 0; i < len; ++i) {
        int index = s[i] - 'a';
        while(p->next[index]==NULL&&p!=root) p = p->fail;
        p = p->next[index];
        p = (p==NULL)?root:p;
        Trie_node *temp = p;
        while(temp != root && temp->counts!=-1) {
            cnt += temp->counts;
            temp->counts = -1;
            temp = temp->fail;
        }
    }
    return cnt;
}
int T,n;
char s[N];
int main() {
    scanf("%d",&T);
    while(T--) {
        Trie_node *root = new Trie_node();
        scanf("%d",&n);
        for(int i = 1; i <= n; ++i)  {
            scanf("%s",s);
            insert(s,root);
        }
        build_ac_automation(root);
        scanf("%s",str);
        printf("%d
",query(str,root));
    }
    return 0;
}