HDU 2222

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

做题还是记录一下比较好 不然 你也不知道自己到底写过什么题目。。
直接上AC自动机即可，还是比较容易理解 bfs保证了最长的相同后缀。构造的并非AC自动机而是trie图...

int T,n,cnt;
char a[MAXN];
int t[MAXN][26],vis[MAXN],fail[MAXN],q[MAXN],mark[MAXN];
inline void insert(int x)
{
	int len=strlen(a+1);
	int p=0;
	rep(1,len,i)
	{
		int c=a[i]-'a';
		if(!t[p][c])
		{
			t[p][c]=++cnt;fail[cnt]=0;mark[cnt]=0;
			memset(t[cnt],0,sizeof(t[cnt]));
		}
		p=t[p][c];
	}
	vis[x]=p;
}
inline void AC()
{
	int l=0,r=0;
	for(int i=0;i<=25;++i)if(t[0][i])q[++r]=t[0][i];
	while(++l<=r)
	{
		int x=q[l];
		for(int i=0;i<=25;++i)
		{
			if(t[x][i])q[++r]=t[x][i],fail[q[r]]=t[fail[x]][i];
			else t[x][i]=t[fail[x]][i];
		}
	}
	int len=strlen(a+1);
	int p=0;
	for(int i=1;i<=len;++i)
	{
		int c=a[i]-'a';
		p=t[p][c];
		mark[p]=1;
	}
	for(int i=r;i>=1;--i)mark[fail[i]]|=mark[i];
}
int main()
{
	freopen("1.in","r",stdin);
	scanf("%d",&T);
	while(T--)
	{
		get(n);cnt=0;memset(t[0],0,sizeof(t[0]));
		rep(1,n,i)scanf("%s",a+1),insert(i);
		scanf("%s",a+1);AC();
		int ans=0;
		for(int i=1;i<=n;++i)if(mark[vis[i]])++ans;
		put(ans);
	}
	return 0;
}