HDU-2222

• Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total
Submission(s):
39064    Accepted
Submission(s): 12596

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

题意：看现在的主字符串，有多少个零碎的字符串构成，
做法:AC自动机模板
代码如下：
#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<cmath>
#include<queue>
#define maxn 500010
using namespace std;
struct Tire
{
    int next[maxn][],fail[maxn],end[maxn];
    int root,L;
    int newnode()
    {
        for(int i=; i<; i++)
        {
            next[L][i] = -;
        }
        end[L++] = ;
        return L-;
    }
    void init()
    {
                    L = ;
                    root = newnode();
    }
    void insert(char buf[])
    {
                    int len = strlen(buf);
                    int now = root;
                    for(int i=;i<len;i++)
                    {
                              if(next[now][buf[i] - 'a'] == -)
                              {
                                        next[now][buf[i] -'a'] = newnode();
                              }
                              now = next[now][buf[i] -'a'];
                    }
                    end[now]++;
    }
    void build()
    {
                    queue<int>Q;
                    fail[root] = root;
                    for(int i=;i<;i++)
                    {
                              if(next[root][i] == -)
                              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=;i<;i++)
                              {
                                       if(next[now][i] == -)
                                       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 = ;
                    for(int i=;i<len;i++)
                    {
                              now = next[now][buf[i] -'a'];
                              int tmp = now;
                              while(tmp != root)
                              {
                                        res += end[tmp];
                                        end[tmp] = ;
                                        tmp = fail[tmp];
                              }
                    }
                    return res;
    }
};
char buf[];
Tire ac;
int main()
{
#ifndef ONLINE_JUDGE
          freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
          int T;
          scanf("%d",&T);
          while(T--)
          {
                    int n;
                    scanf("%d",&n);
                    ac.init();
                    for(int i=;i<n;i++)
                    {
                              scanf("%s",buf);
                              ac.insert(buf);
                    }
                    ac.build();
                    scanf("%s",buf);
                    printf("%d\n",ac.query(buf));
          }
          return ;
}