Hdu-2222 Keywords Search（AC自动机）

Problem Description

In the modern time, Search engine came into the life of everybody like Google, Baidu, etc.
Wiskey(威士忌酒) also wants to bring this feature(特写) to his image retrieval(检索) 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(简化) 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(整数) 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

 

Recommend

lcy

 

分析：AC自动机模板题。

#include <queue>
#include <cstdio>
#include <cstring> 
#include <iostream>
#define MAXNODE 500005
using namespace std;
int n,T;
char s[1000005];
struct ACautomata
{
	int ch[MAXNODE][26];
	int f[MAXNODE];             // fail函数
	int val[MAXNODE]; 	        // 是否为单词结尾 
	int last[MAXNODE];          // 后缀链接 
	int cnt[10005];             // 每个单词出现次数 
	int tot;                    // trie 单词总数 
	int num;                    // 单词出现了几个 
	int time[10005];
	void init()
	{
		num = 0;
		tot = 1;
		memset(ch[0],0,sizeof(ch[0]));
		memset(cnt,0,sizeof(cnt));
		memset(time,0,sizeof(time));
	}
	int idx(char c)
	{
		return c - 'a';
	}
	void insert(char *s,int v)
	{
		int u = 0,n = strlen(s);
		for(int i = 0;i < n;i++)
		{
			int c = idx(s[i]);
			if(!ch[u][c])
			{
				memset(ch[tot],0,sizeof(ch[tot]));
				val[tot] = 0;
				ch[u][c] = tot++; 
			}
			u = ch[u][c];
		}
		if(val[u]) time[val[u]]++;
		else val[u] = v,time[v] = 1;
	}
	void print(int i,int j)
	{
		if(j) 
		{
			if(!cnt[val[j]]) num += time[val[j]];
			cnt[val[j]]++;
			print(i,last[j]);
		}
	}
	void find(char *T)
	{
		int n = strlen(T);
		int j = 0;
		for(int i = 0;i < n;i++)
		{
			int c = idx(T[i]);
			j = ch[j][c];
			if(val[j]) print(i,j);
			else if(last[j]) print(i,last[j]);
		}
	}
	void getFail()
	{
		queue<int> q;
		f[0] = 0;
		for(int c = 0;c < 26;c++)
		{
			int u = ch[0][c];
			if(u)
			{
				f[u] = 0;
				q.push(u);
				last[u] = 0;
			}
		}
		while(!q.empty())
		{
			int r = q.front();q.pop();
			for(int c = 0;c < 26;c++)
			{
				int u = ch[r][c];
				if(!u)
				{
					ch[r][c] = ch[f[r]][c];
					continue;
				}
				q.push(u);
				int v = f[r];
				f[u] = ch[v][c];
				last[u] = val[f[u]] ? f[u] : last[f[u]];
			}
		}
	}
} tree;
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		tree.init();
		scanf("%d",&n);
		for(int i = 1;i <= n;i++)
		{
			scanf("%s",s);
			tree.insert(s,i);
		}
		tree.getFail();
		scanf("%s",s);
		tree.find(s);
		cout<<tree.num<<endl;
	}
}