hdu 2846 Repository - 字典树

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 字典树变式
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <iostream>
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] != '\0')
	{
		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 做标记 增加数量
		{
			t->count ++;
			t->no = noo;
		}
		i ++;
	}

}

int func(Trie* trie,char s[])
{
	Trie* t = trie;
	Trie* tpre;
	int i = 0;
	while(s[i] != '\0')
	{
		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);
		/*
			这里对于stmp = "abc" 分为 abc，bc，c这3个字符串插入 
				一次插入后如下图
					   root
					  / |  \
					a   b   c 
				   /    |
				  b     c
				 /
				c
			每次都这样处理， 相应字符的count 会加上去，最后统计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\n" , func(trie,stmp) ) ;
	}
	return 0;
}