HDOJ 题目2846 Repository（字典树）

最新推荐文章于 2018-07-12 20:02:54 发布

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本文深入探讨了信息技术领域的多个细分技术领域，包括前端开发、后端开发、移动开发、游戏开发等，涵盖了从基础到进阶的技术知识，为读者提供了一个全面的视角来了解和掌握信息技术的最新动态。

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Repository

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2656 Accepted Submission(s): 1040

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

Source

2009 Multi-University Training Contest 4 - Host by HDU

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ac代码

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
typedef struct s
{
	struct s * child[26];
	int cnt,id;
}node,*Node;
Node root;
void insert(char *s,int id)
{
	Node cur,newnode;
	int i,now,len;
	len=strlen(s);
	cur=root;
	for(i=0;i<len;i++)
	{
		now=s[i]-'a';
		if(cur->child[now]!=NULL)
		{
			cur=cur->child[now];
		}
		else
		{
			newnode=(Node)calloc(1,sizeof(node));
			cur->child[now]=newnode;
			cur=cur->child[now];
			cur->cnt=0;
			cur->id=-1;
		}
		if(cur->id!=id)
		{
			cur->cnt++;
			cur->id=id;
		}
	}
}
void seach(char *s)
{
	Node cur;
	cur=root;
	int i,len=strlen(s);
	for(i=0;i<len;i++)
	{
		int now=s[i]-'a';
		if(cur->child[now]==NULL)
			break;
		cur=cur->child[now];
	}
	if(i<len)
		printf("0\n");
	else
		printf("%d\n",cur->cnt);
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int i,j;
		char str[21];
		root=(Node)calloc(1,sizeof(node));
		for(i=0;i<n;i++)
		{
			scanf("%s",str);
			int len=strlen(str);
			for(j=0;j<len;j++)
				insert(str+j,i);
		}
		int m;
		scanf("%d",&m);
		while(m--)
		{
			scanf("%s",str);
			seach(str);
		}
	}
}