hdu 2846 Repository - 字典樹

Repository

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3247 Accepted Submission(s): 1227


Problem Description When you go shopping, you can search in repository for avalible merchandises by the computers and internet. First you give the search system a name about something, then the system responds with the results. Now you are given a lot merchandise names in repository and some queries, and required to simulate the process.
Input There is only one case. First there is an integer P (1<=P<=10000)representing the number of the merchanidse names in the repository. The next P lines each contain a string (it's length isn't beyond 20,and all the letters are lowercase).Then there is an integer Q(1<=Q<=100000) representing the number of the queries. The next Q lines each contains a string(the same limitation as foregoing descriptions) as the searching condition.
Output For each query, you just output the number of the merchandises, whose names contain the search string as their substrings.
Sample Input

20
ad
ae
af
ag
ah
ai
aj
ak
al
ads
add
ade
adf
adg
adh
adi
adj
adk
adl
aes
5
b
a
d
ad
s

Sample Output

0
20
11
11
2

Source 2009 Multi-University Training Contest 4 - Host by HDU

/*
http://acm.hdu.edu.cn/showproblem.php?pid=2846
Repository 字典樹變式
*/
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef struct node{
	int no;
	int count;
	struct node* next[27];
	node(int _count = 0)
	{
		count = _count;
		no = -1;
		int i;
		for(i = 0 ; i < 27 ; i ++)
		{
			next[i] = NULL;
		}
	}
}Trie;

void insertNode(Trie* trie , char* s,int noo)
{
	Trie* t = trie;
	int i = 0;
	while(s[i] != '')
	{
		int tmp = s[i]-'a';
		if(t->next[tmp] == NULL)
		{
			t->next[tmp] = new node(0);
		}
		t = t->next[tmp];
		if(t->no != noo) // noo 做標記 增加數量
		{
			t->count ++;
			t->no = noo;
		}
		i ++;
	}

}

int func(Trie* trie,char s[])
{
	Trie* t = trie;
	Trie* tpre;
	int i = 0;
	while(s[i] != '')
	{
		int tmp = s[i]-'a';
		if(t->next[tmp] == NULL)
		{
			return 0;
		}
		tpre = t;
		t = t->next[tmp];
		i ++;
	}
	return t->count;
}

int main()
{
	//freopen(in.txt,r,stdin);
	int n , m ;
	int i , j ;
	scanf(%d,&n);
	char stmp[21];
	Trie* trie = new node(0);
	for(i = 0 ; i < n ; i ++)
	{
		scanf(%s,stmp);
		int len = strlen(stmp);
		/*
			這裡對於stmp = abc 分為 abc，bc，c這3個字符串插入 
				一次插入後如下圖
					   root
					  / |  
					a   b   c 
				   /    |
				  b     c
				 /
				c
			每次都這樣處理， 相應字符的count 會加上去，最後統計count就行了
		*/
		for(j = 0;j < len ;j ++)
		{
			insertNode(trie , stmp+j , i);
		}
	}

	scanf(%d , &m) ;
	for(i = 0 ; i < m ; i ++)
	{
		scanf(%s,stmp);
		printf(%d
 , func(trie,stmp) ) ;
	}
	return 0;
}