HDU2846 Repository 字典树

Repository

                                                                                                           Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                                                                 Total Submission(s): 961    Accepted Submission(s): 313

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

　　子串，听到这个就头麻，确实这个问题比较纠结，YY通过转换成数字来相除，徒劳啊。。。

暴力方法是将所给长为N单词拆成N个单词，然后就是基本的查找的，这里要注意的就是将同一个单词中重复的部分忽略掉。利用结构题中的新增变量来记录一个节点上一次覆盖的单词。

   代码如下：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct Node
{
	int flag;
	int pos;
	struct Node *child[26];
}Node;

Node *init(  )
{
	Node *n= ( Node * )malloc( sizeof( Node ) );
	n-> flag= 0;
	n-> pos= -1;
	memset( n-> child, NULL, sizeof( n-> child ) );
	return n;
}

void insert( Node *p, char *in, int pos )
{
	if( *in== '\0' )
	{
		return;
	}
	else
	{
		if( p-> child[ *in- 'a' ]== NULL )
		{
			p-> child[ *in- 'a' ]= init(  );
		}
		if( p-> child[ *in- 'a' ]-> pos!= pos )
		{
		    p-> child[ *in- 'a' ]->flag++;
		}
		p-> child[ *in- 'a' ]-> pos= pos;
		insert( p-> child[ *in- 'a' ], in+ 1, pos );
	}
}

int search( Node *p, char *in )
{
	if( *in== '\0' )
	{
		if( p-> flag )
		{
			return p-> flag; 
		}
		else
		{
			return 0;
		}
	}
	else
	{
		if( p-> child[ *in- 'a' ]== NULL )
		{
			return 0;
		}
		search( p-> child[ *in- 'a' ], in+ 1 );
	}
}

int main(  )
{
	int p, q, len;	char in[25];	Node *n= init(  );
	scanf( "%d", &p );
	while( p-- )
	{
		scanf( "%s", in );
		len= strlen( in ); 
		for( int i= 0; i< len; ++i )
		{
			insert( n, in+ i, p );
		}
	}
	scanf( "%d", &q );
	while( q-- )
	{
		scanf( "%s", in );
		printf( "%d\n", search( n, in ) );
	}
}