HDU - 2222 Keywords Search（AC自动机模板题）

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本文介绍了一个关键词搜索算法的实现过程，包括使用AC自动机进行高效关键词匹配的方法。该算法能够处理大量关键词，并在文本中快速查找匹配项。文章详细解释了如何构建自动机、进行模式匹配及优化技巧。

Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 67795 Accepted Submission(s): 22825

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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Statistic | Submit | Discuss | Note

#include <bits/stdc++.h>
using namespace std;

int t, n;
char s[101], f[1000001];
struct xx{
    int cnt;
    xx *fail;
    xx *nex[26];
    xx(){
        memset(nex, NULL, sizeof(nex));
        fail = NULL;
        cnt = 0;
    }
};

void Insert(xx *root, char a[]){
    int l = strlen(a);
    xx *p = root;
    for(int i = 0; i < l; i++){
        int id = a[i]-'a';
        if(p->nex[id] == NULL){
            p->nex[id] = new xx();
        }
        p = p->nex[id];
    }
    (p->cnt)++;
}

void bfs(xx *root){
    queue<xx*> k;
    root->fail = NULL;
    k.push(root);
    xx *p, *tmp;
    while(!k.empty()){
        tmp = k.front();
        k.pop();
        for(int i = 0; i < 26; i++){
            if(tmp->nex[i] != NULL){
                p = tmp->fail;
                while(p && !p->nex[i]){
                    p = p->fail;
                }
                if(!p) tmp->nex[i]->fail = root;
                else tmp->nex[i]->fail = p->nex[i];
                k.push(tmp->nex[i]);
            }
        }
    }
}

int AC_run(xx *root, char a[]){
    int l = strlen(a);
    int ans = 0;
    xx *p = root;
    for(int i = 0; i < l; i++){
        int id = a[i]-'a';
        while(!p->nex[id] && p != root){
            p = p->fail;
        }
        p = p->nex[id];
        if(!p) p = root;
        xx *tmp = p;
        while(tmp != root && tmp->cnt != -1){
            ans += tmp->cnt;
            tmp->cnt = -1;
            tmp = tmp->fail;
        }
    }
    return ans;
}

int main(){
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        xx *root = new xx();
        for(int i = 0; i < n; i++){
            scanf("%s", s);
            Insert(root, s);
        }
        bfs(root);
        scanf("%s", f);
        printf("%d\n", AC_run(root, f));
    }
}