HDU2222 -ac自动机

Keywords Search

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 63350    Accepted Submission(s): 21015


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

 

题目大意：

给n 个字母，以及一个串，问这个串中有多少个字母，不能重复计数

题目思路：

AC自动机模板，对于AC自动机的讲解可以看这篇博文：ac自动机详解

AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;

const int maxn = 1e6+10;
const int maxtot = 5e5+10;
int Size;

struct Tree
{
int nex[26];
int fail,cnt;
void init()
{
fail = cnt = 0;
memset(nex,0,sizeof(nex));
}
}node[maxtot];

void Insert(char *s)
{
int n = strlen(s);
int now = 0;
for(int i=0;i<n;i++)
{
int id = s[i]-'a';
if(!node[now].nex[id])
{
node[now].nex[id] = Size++;
}
now = node[now].nex[id];
}
node[now].cnt++;
}
queue<int>q;
void Build()
{
node[0].fail = -1;
q.push(0);
while(!q.empty())
{
int u = q.front();
q.pop();
for(int i=0;i<26;i++)
{
if(node[u].nex[i])
{
if(u==0)node[node[u].nex[i]].fail = 0;
else
{
int v = node[u].fail ;
while(v!=-1)
{
if(node[v].nex[i])
{
node[node[u].nex[i]].fail = node[v].nex[i];
break;
}
v = node[v].fail;
}
if(v==-1)node[node[u].nex[i]].fail = 0;
}
q.push(node[u].nex[i]);
}
}
}
}

int Search(char *s)
{
int n = strlen(s);
int res = 0,now = 0;
for(int i=0;i<n;i++)
{
int id = s[i]-'a';
if(node[now].nex[id])now = node[now].nex[id];
else
{
int p = node[now].fail;
while(p!=-1&&node[p].nex[id]==0)p = node[p].fail;
if(p==-1)now = 0;
else now = node[p].nex[id];
}

if(node[now].cnt)
{
int uu = now;
while(uu)
{
res+=node[uu].cnt;
node[uu].cnt = 0;
uu = node[uu].fail;
}
}

}

return res;
}

void init()
{

for(int i=0;i<maxtot;i++)
node[i].init();
Size = 1;
}
char s[maxn];
int main()
{
int t;cin>>t;
while(t--)
{
int n;
scanf("%d",&n);
init();
for(int i=0;i<n;i++)
{
scanf("%s",s);
Insert(s);
}
Build();
scanf("%s",s);
printf("%d\n",Search(s));

}
return 0;
}