hdu2222(AC自动机模板题)

感觉最近撸数据结构收获不少~

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2222

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 74272    Accepted Submission(s): 25538

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

 

Author

Wiskey

思路：第一道AC自动题（虽然只是一道最简单的模板题。。。）通过这道题，我对Trie的用法和写法更加的熟悉了（。。。），感觉AC自动机的确和KMP很像，比较难的地方就是fail指针的构造，还得细究。

上代码：

#include
#include
#include
#include
using namespace std;
struct Trie
{
    int next[500010][26], fail[500010], end[500010];
    int root, L;
    int newnode()
    {
        for (int i = 0; i < 26; i++)
        {
            next[L][i] = -1;
        }
        end[L++] = 0;
        return L - 1;
    }

    void init()
    {
        L = 0;
        root = newnode();
    }

    void insert(char buf[])
    {
        int len = (int)strlen(buf);
        int now = root;
        for (int i = 0; i < len; i++)
        {
            if (next[now][buf[i] - 'a'] == -1)
            {
                next[now][buf[i] - 'a'] = newnode();
            }
            now = next[now][buf[i] - 'a'];
        }
        end[now]++;
    }

    void build()
    {
        queueQ;
        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 = (int)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[1000010];
Trie ac;
int main()
{
    int t;
    int 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;
}