Hdu2222 Keywords Search

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 70386    Accepted Submission(s): 23963

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

 

Recommend

lcy

题目大意：给定n个单词和一篇文章，问有多少个单词出现在文章里了.

分析：AC自动机模板题.多模式匹配问题.

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

using namespace std;

const int maxn = 10010 * 55;

int T, n, tot = 1, vis[maxn], ans;
char s[maxn * 2];

struct node
{
    int tr[30], cnt, fail;
    void clear()
    {
        memset(tr, 0, sizeof(tr));
        cnt = 0;
        fail = 0;
    }
}e[maxn];

void insert(char *ss)
{
    int len = strlen(ss + 1);
    int u = 1;
    for (int i = 1; i <= len; i++)
    {
        int ch = ss[i] - 'a';
        if (!e[u].tr[ch])
        {
            e[u].tr[ch] = ++tot;
            e[tot].clear();
        }
        u = e[u].tr[ch];
    }
    e[u].cnt++;
}

void build()
{
    for (int i = 0; i < 26; i++) //易错
        e[0].tr[i] = 1;
    queue <int> q;
    q.push(1);
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        int fail = e[u].fail;
        for (int i = 0; i < 26; i++)
        {
            int y = e[u].tr[i];
            if (y)
            {
                e[y].fail = e[fail].tr[i];
                q.push(y);
            }
            else
                e[u].tr[i] = e[fail].tr[i];
        }
    }
    /*
    for (int i = 1; i <= tot; i++)
        cout << i << " " << e[i].fail << endl;
        */
}

void solve()
{
    scanf("%s", s + 1);
    int len = strlen(s + 1),u = 1;
    for (int i = 1; i <= len; i++)
    {
        int p = s[i] - 'a';
        while (u && !e[u].tr[p])
            u = e[u].fail;
        u = e[u].tr[p];
        int temp = u;
        //cout << u << endl;
        while (temp && e[temp].cnt != -1)
        {
            ans += e[temp].cnt;
            e[temp].cnt = -1;
            temp = e[temp].fail;
        }
    }
}

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        ans = 0;
        e[tot = 1].clear();
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
        {
            scanf("%s", s + 1);
            insert(s);
        }
        build();
        solve();
        printf("%d
", ans);
    }

    return 0;
}