HDU2222 Keywords Search(AC自动机)

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 68126    Accepted Submission(s): 22973

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

 

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题目的意思是给出若干个字符串，问有多少个是给定串的子串

思路：ac自动机模板题

#include<bits/stdc++.h>

using namespace std;
const int maxn=500005;

struct Trie
{
    int next[maxn][26],fail[maxn],end[maxn];
    int root,L;
    int newnode()
    {
        memset(next[L],-1,sizeof next[L]);
        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]++;
    }

    bool findintree(char buf[])
    {
        int len =strlen(buf);
        int now=root;
        int flag=0;
        for(int i=0; i<len; i++)
        {
            if(next[now][buf[i]-'a']!=-1)
            {
                flag=1;
                break;
            }
            now=next[now][buf[i]-'a'];
        }
         return end[now]&&!flag;
    }



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;
}
void debug()
{
    printf("id =    ,fail =    ,end =    ,chi = [ a b c d e f g h i j k l m n o p q r s t u v w x y z]\n");
    for(int i = 0; i < L; i++)
    {
        printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);
        for(int j = 0; j < 26; j++)
            printf("%2d",next[i][j]);
        printf("]\n");
    }
}
};
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;
}