Hdu 2222 . Keywords Search

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

Solution

• 这是一道经典的多模式串匹配问题，也是一道AC自动机模板题。

• 不多说，上代码：

Code

#include<cstdio>#include<cstring>#include<queue>using namespace std;const int N=500005;struct AC_Automation{    int siz,ans;    int trie[N][26],v[N],next[N],fail[N];    void init()    {        siz=ans=0;        memset(v,0,sizeof(v));        memset(fail,0,sizeof(fail));        memset(trie[0],0,sizeof(trie[0]));    }    void insert(char *p)    {        int len=strlen(p),j=0;        for(int i=0;i<len;i++)        {            if(!trie[j][p[i]-'a'])            {                trie[j][p[i]-'a']=++siz;                memset(trie[siz],0,sizeof(trie[siz]));            }            j=trie[j][p[i]-'a'];        }        v[j]++;    }    void getfail()    {        queue<int>que;        for(int i=0;i<26;i++)        {            int x=trie[0][i];            if(x)            {                fail[x]=next[x]=0;                que.push(x);                    }        }        while(!que.empty())        {            int now=que.front();            que.pop();            for(int i=0;i<26;i++)            {                int x=trie[now][i],y=fail[now];                if(!x)                {                    trie[now][i]=trie[y][i];                    continue;                }                que.push(x);                while(y && !trie[y][i]) y=fail[y];                fail[x]=trie[y][i];            }            next[now]=v[fail[now]]?fail[now]:next[fail[now]];        }    }    void find(char *p)    {        int len=strlen(p);        getfail();        for(int i=0,j=0;i<len;i++)        {            j=trie[j][p[i]-'a'];            for(int k=j;k;k=next[k])                if(v[k]) ans+=v[k],v[k]=0;        }    }}AC;int main(){    int T,n;    char p[55],s[N<<1];    scanf("%d",&T);    while(T--)    {        AC.init();        scanf("%d",&n);        while(n--)        {            scanf("%s",&p);            AC.insert(p);        }        scanf("%s",&s);        AC.find(s);        printf("%d\n",AC.ans);    }    return 0;}