hdu 2222(Keywords Search) (AC自动机)

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 34147    Accepted Submission(s): 11061


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

 

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【题意】有几个单词在目标串中出现
【思路】估计是最入门的AC自动机吧，不过网上的代码有的是错的，虽然能ac,呵呵。。
discuss里的反例

12
6
a
ba
cba
dcba
baf
f
dcbafd

【代码】

/*
kmp+tire
*/
#include<cstdio>
#include<cstring>
#include<queue>
#include<map>
#include<iostream>
using namespace std;

int const MAX_NODE=52*10100;
int const MAX_SIZE=26;

int tot;
int ch[MAX_NODE][MAX_SIZE];
int value[MAX_NODE];
//map<int,string>II;

void Build_Trie(char *word,int key)
{
int u=0;
for(int i=0;word[i];i++)
{
int v=word[i]-'a';
if(!ch[u][v])
{
memset(ch[tot],0,sizeof(ch[tot]));
value[tot]=0;
ch[u][v]=tot++;
}
u=ch[u][v];
}
value[u]+=key;
//II[u]=(string)word;
}

void print(int root,char *s,int len)
{
if(value[root])
printf("%s\n",s);
for(int i=0;i<MAX_SIZE;i++)
{
s[len]=i+'a';s[len+1]=0;
if(ch[root][i])
print(ch[root][i],s,len+1);
}
}

int Fail[MAX_NODE];
int Ac(char *T)
{
int ans=0;
int len=strlen(T);
int j=0;
for(int i=0;i<len;i++)
{
int u=T[i]-'a';
while(j&&!ch[j][u])j=Fail[j];
j=ch[j][u];
//cout<<"Fail j="<<j<<endl;
int temp=j;
while(temp)//这里可以改写成while(temp&&value[temp]!=-1)
{
//printf("%d\n",value[j]);
if(value[temp])
{
//cout<<II[temp]<<endl;
ans+=value[temp];
value[temp]=0;//按上面的,就需要写成value[temp]=-1;
}
temp=Fail[temp];
}
}
return ans;
}

queue<int>Q;
void GetFail()
{
Fail[0]=0;
for(int c=0;c<MAX_SIZE;c++)
{
int u=ch[0][c];
if(u)
{
Fail[u]=0;
Q.push(u);
//last
}
}
while(!Q.empty())
{
int r=Q.front();Q.pop();
for(int c=0;c<MAX_SIZE;c++)
{
int u=ch[r][c];
if(!u)continue;
Q.push(u);
int v=Fail[r];
while(v&&!ch[v][c])
v=Fail[v];
Fail[u]=ch[v][c];
//last[u]=value[Fail[u]]?Fail[u]:last[Fail[u]];
}
}

}
void innit()
{
memset(ch[0],0,sizeof(ch[0]));
tot=1;
value[0]=0;
}
char T[1000005];
int main()
{
int Cas;
scanf("%d",&Cas);
while(Cas--)
{
int n;
scanf("%d",&n);
char s[100];
innit();
while(n--)
{
// cout<<"n: "<<n<<endl;
scanf("%s",s);
Build_Trie(s,1);
}
//s[0]=0;
//print(0,s,0);
//printf("\n");
GetFail();
scanf("%s",T);
int ans=Ac(T);
printf("%d\n",ans);
}
return 0;
}