Keywords Search HDU - 2222 （ ac自动机）模版题

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 77903    Accepted Submission(s): 27032

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：给你几个模式串， 再给你一个长串，问那几个模式串在长串当中一共出现了几次

思路：其实就是ac自动机的模版

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<queue>
#include<set>
#include<vector>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
#define ll long long
const int maxn=1000;
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=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(){
        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[1000010];
Trie ac;
int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        int n;
        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;
}