Repository
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1129 Accepted Submission(s): 382
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
#include <iostream>
#include<cstring>
#define MAX 26
using namespace std;
typedef struct Trie_Node
{
int count; //记录包含该结点的单词个数
int id; //最后一次经过此结点的商品ID
Trie_Node *next[MAX];
}Trie;
void insert(Trie *root,char *s,int id)
{
Trie *p=root;
while(*s!='\0')
{
if(p->next[*s-'a']==NULL)
{
Trie *temp=(Trie *)malloc(sizeof(Trie));
for(int i=0;i<MAX;i++)
{
temp->next[i]=NULL;
}
temp->count=0;
temp->id=-1; //-1表示没有商品
p->next[*s-'a']=temp;
}
p=p->next[*s-'a'];
if(p->id!=id) //如果当前结点的商品ID不等于要插入商品的ID,则计数器count++,并且重新置ID的值
{
p->id=id;
p->count++;
}
s++;
}
}
int search(Trie *root,char *s)
{
Trie *p=root;
for(int i=0;s[i]!='\0';i++)
{
if(p->next[s[i]-'a']==NULL)
return 0;
p=p->next[s[i]-'a'];
}
return p->count;
}
int main(int argc, char *argv[])
{
int i,j;
int n,m;
char s[21];
Trie *root=(Trie *)malloc(sizeof(Trie));
for(i=0;i<MAX;i++)
{
root->next[i]=NULL;
}
root->count=0;
root->id=-1;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%s",s);
for(j=0;j<strlen(s);j++) //将字符串X=X1X2...Xn的分别以X1,X2...Xn开头的后缀字符串插入到Trie树中
{
insert(root,s+j,i);
}
}
scanf("%d",&m);
for(i=0;i<m;i++)
{
scanf("%s",s);
printf("%d\n",search(root,s));
}
return 0;
}

本文介绍了一种利用Trie树进行高效字符串搜索的方法,通过特殊构造使得能够判断一个字符串是否作为另一个字符串的子串出现。文章详细解释了如何构建Trie树并实现字符串的插入与查询过程。
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