AC自动机实现多模式字符串匹配-优快云博客

本文链接：https://blog.youkuaiyun.com/computer_user/article/details/77670288

本文介绍了一种利用AC自动机实现的多模式字符串匹配算法，通过构建自动机来高效地查找文本中多个关键词的位置。文章提供了一个完整的代码示例，包括自动机的建立过程、关键词插入方法及查询流程。

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Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 67091 Accepted Submission(s): 22567

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

Author

Wiskey

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网上很多讲解AC自动机的算法了，我就不说了。

一开始听到这名字以为是什么逆天的算法，能帮我把任何题目都AC掉。。。

emmmmm。。好吧，只是个多模式的字符串匹配算法。。

废话少说，贴模板。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int maxn = 1e6+5;
const int maxs = 55;
char s[maxs],str[maxn];
int i,j,k,n;

struct node {
    node *fail;
    node *next[26];
    int cnt;
    node() {
       fail = NULL;
       cnt = 0;
       for (int k=0; k<26; k++) next[k] = NULL;
    }
}*root;
queue<node *>que;

void Insert(){
     node *p = root;
     int len = strlen(s);
     int tmp;
     for (i=0; i<len; i++) {
         tmp = s[i] - 'a';
         if (p->next[tmp] == NULL) p->next[tmp] = new node();
         p = p->next[tmp];

     }
     p->cnt++;
}

void build_ac() {
     que.push(root);
     while (!que.empty()) {
        node *p = que.front();
        que.pop();
        for (i=0; i<26; i++) if (p->next[i]!=NULL) {
             if (p==root) p->next[i]->fail = root;
             else {
                 node *temp = p->fail;
                 while (temp!=NULL) {
                    if (temp->next[i] !=NULL) {
                        p->next[i]->fail = temp->next[i];
                        break;
                    }
                    temp = temp->fail;
                 }

                 if (temp==NULL) p->next[i]->fail = root;
             }
             que.push(p->next[i]);
        }
     }
}

int query(){
   int ans = 0,index;
   node *p = root, *temp;
   int len = strlen(str);
   for (i=0; i<len; i++) {
       index = str[i] - 'a';
       while (p->next[index]==NULL && p!=root) p = p->fail;
       p = p->next[index];
       if (p==NULL) p = root;
       temp = p;
       while (temp !=root && temp->cnt!=-1) {
          ans += temp->cnt;
          temp->cnt = -1;
          temp = temp->fail;
       }
   }
   return ans;
}

int main(){
    int T;
    scanf("%d",&T);
    while (T--) {
        scanf("%d",&n);
        root = new node();
        for (k=1; k<=n; k++) {
            scanf("%s",s);
            Insert();
        }
        build_ac();
        scanf("%s",str);
        printf("%d\n",query());
    }
    return 0;
}