HDU 2222 Keywords Search（AC自动机/模板题）

最新推荐文章于 2020-02-25 16:26:17 发布

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最新推荐文章于 2020-02-25 16:26:17 发布

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文章标签： java

原文链接：http://www.cnblogs.com/ZhaoxiCheung/p/7195947.html

本文介绍了一种高效的关键词搜索算法实现，该算法使用了Trie树和Aho-Corasick自动机来快速匹配大量关键词，适用于图像检索系统的场景。通过构建Trie树并进行优化，能够实现在长文本中高效查找关键词。

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Keywords Search

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 63941 Accepted Submission(s): 21264

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

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

struct Trie{
    int cnt;
    Trie* next[26];
    Trie* behind;
};


int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
    {
        int N;
        char str[60];
        Trie* root = new Trie;
        root->cnt = -1;
        root->behind = NULL;
        for (int i = 0;i < 26;i++)  root->next[i] = NULL;
        scanf("%d",&N);
        while (N--)
        {
            scanf("%s",str);
            Trie* p = root;
            int i = 0;
            while (str[i])
            {
                int j = str[i] - 'a';
                if (!p->next[j])
                {
                    Trie* q = new Trie;
                    q->cnt = 0;
                    q->behind = NULL;
                    for (int k = 0;k < 26;k++)  q->next[k] = NULL;
                    p->next[j] = q;
                }
                p = p->next[j];
                i++;
            }
            p->cnt++;
        }
        queue<Trie*>que;
        root->behind = root;
        for (int i = 0;i < 26;i++)
        {
            if (!root->next[i])
            {
                root->next[i] = root;
            }
            else
            {
                root->next[i]->behind = root;
                que.push(root->next[i]);
            }
        }
        while (!que.empty())
        {
            Trie* p = que.front();
            Trie* q = p->behind;
            que.pop();
            for (int i = 0;i < 26;i++)
            {
                if (!p->next[i])
                {
                    p->next[i] = q->next[i];
                }
                else
                {
                    p->next[i]->behind = q->next[i];
                    que.push(p->next[i]);
                }
            }
        }
        char article[1000005];
        Trie* p = root;
        scanf("%s",article);
        int res = 0;
        int i = 0;
        while (article[i])
        {
            int j = article[i] - 'a';
            p = p->next[j];
            for (Trie* tmp = p;tmp != root && tmp->cnt !=-1;tmp = tmp->behind)
            {
            	res += tmp->cnt;
            	tmp->cnt = -1;    //计算完毕，置-1表示已经计算过
            }
            i++;
        }
        printf("%d\n",res);
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N = 500005;

struct AC_Automan{
    int next[N][26];
    int fail[N];
    int cnt[N];
    int root,tot;
    int newnode()
    {
        for (int i = 0;i < 26;i++)  next[tot][i] = -1;
            fail[tot] = -1;
        cnt[tot] = 0;
        return tot++;
    }
    void clear()
    {
        tot = 0;
        root = newnode();
    }
    void insert(const string &s)
    {
        int p = root;
        for (int i = 0,len = s.length();i < len;i++)
        {
            if (next[p][s[i]-'a'] == -1)    next[p][s[i]-'a'] = newnode();
            p = next[p][s[i]-'a'];
        }
        cnt[p]++;
    }
    void build()
    {
        queue<int>Q;
        Q.push(root);
        while (!Q.empty())
        {
            int p = Q.front();Q.pop();
            for (int i = 0;i < 26;i++)
            {
                if (~next[p][i])
                {
                    if (p == root)  fail[next[p][i]] = root;
                    else    fail[next[p][i]] = next[fail[p]][i];
                    Q.push(next[p][i]);
                }
                else
                {
                    if (p == root)  next[p][i] = root;
                    else    next[p][i] = next[fail[p]][i];
                }
            }
        }
    }
    int solve(const string &t)
    {
        int p = root,ans = 0;
        for (int i = 0,len = t.length();i < len;i++)
        {
            p = next[p][t[i]-'a'];
            for (int tmp = p;tmp != root && cnt[tmp] != -1;tmp = fail[tmp])
            {
                ans += cnt[tmp];
                cnt[tmp] = -1;
            }
        }
        return ans;
    }
}ac;

int main()
{
    cin.sync_with_stdio(0);
    int T;
    cin >> T;
    while (T--)
    {
        ac.clear();
        int n;
        cin >> n;
        for (int i = 0;i < n;i++)
        {
            string s;
            cin >> s;
            ac.insert(s);
        }
        ac.build();
        string t;
        cin >> t;
        cout << ac.solve(t) << endl;
    }
    return 0; 
}

转载于:https://www.cnblogs.com/ZhaoxiCheung/p/7195947.html