【模板练习——AC自动机】Keywords Search HDU - 2222

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本文介绍了一个高效的关键词搜索算法实现，该算法使用AC自动机（Aho-Corasick Automaton），能够在大量文本数据中快速匹配多个关键词。通过构建Trie树并进行预处理，算法能够高效地处理关键词匹配任务。

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1题意：询问在目标串中出现几个模式串

Keywords Search HDU - 2222

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

以下为Accepted代码

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int pat_len = 501400;/*why???*/
const int tar_len = 1001400;

struct Trie{
    int next[pat_len][26], fail[pat_len], end[pat_len];
    int root, L;
    int new_node(){
        for(int i = 0; i < 26; i++)
            next[L][i] = -1;
        end[L++] = 0;
        return L-1;
    }
    void Init(){
        L = 0;
        root = new_node();
    }
    void Insert(char buf[]){
        int len = strlen(buf);
        int now = root;
        for(int i = 0; i < len; i++){
            if(next[now][buf[i]-'a'] == -1){
                next[now][buf[i]-'a'] = new_node();
            }
            now = next[now][buf[i]-'a'];
        }
        end[now]++;
    }
    void build(){
        queue <int> Q;
        fail[root] = root;
        for(int i = 0; i < 26; i++){
            if(next[root][i] == -1)
                next[root][i] = root;
            else {
                fail[next[root][i]] = root;
                Q.push(next[root][i]);
            }
        }
        while(!Q.empty()){
            int now = Q.front();
            Q.pop();
            for(int i = 0; i < 26; i++){
                if(next[now][i] == -1)
                    next[now][i] = next[fail[now]][i];
                else {
                    fail[next[now][i]] = next[fail[now]][i];
                    Q.push(next[now][i]);
                }
            }
        }
    }
    int query(char buf[]){
        int len = strlen(buf);
        int now = root;
        int res = 0;
        for(int i = 0; i < len; i++){
            now = next[now][buf[i]-'a'];
            int temp = now;
            while(temp != root){
                res += end[temp];
                end[temp] = 0;
                temp = fail[temp];
            }
        }
        return res;
    }
};

char buf[tar_len];
Trie ac;

int main(){
    int T, n;
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        ac.Init();
        for(int i = 0; i < n; i++){
            scanf("%s", buf);
            ac.Insert(buf);
        }
        ac.build();
        scanf("%s", buf);
        printf("%d\n", ac.query(buf));
    }
    return 0;
}