HDU 2846Repository（字典树+计数）_字典树计数-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_37164003/article/details/77248032

本文介绍了一种利用字典树进行高效商品搜索的方法。通过构建字典树存储商品名称，并支持模糊搜索功能，文章详细展示了如何插入、查找商品名并计算匹配数量。此外，还提供了一个完整的代码示例。

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Repository

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/65536K (Java/Other)

Total Submission(s) : 17 Accepted Submission(s) : 5

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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

想法：

在字典树上计数，注意防止多个重复计数

代码：

#include<stdio.h>
#include<string.h>
using namespace std;
const int N=26;
struct Trie
{
    int flag;
    int num;
    Trie *nx[N];
};
Trie *root;
void init (Trie *t)
{
    for (int j = 0; j < N; ++j) //初始化
    {
        t->nx[j] = NULL;
    }
     t->num = 0;
    t->flag = -1;
}
void Insert (char str[],int Id)//插入单词
{
    Trie *p = root;
    int len = strlen (str);
    for (int i = 0; i < len; ++i)
    {
        int id = str[i] - 'a';
        if (p->nx[id] == NULL)
        {
            Trie *t = new Trie;
            init(t);
            p->nx[id] = t;
        }
        p=p->nx[id];
        if(p->flag!=Id)
        {
        p->flag=Id;
        p->num++;
        }

    }
}
int Find(char str[])
{
    Trie *p = root;
    int len = strlen(str);
    for (int i = 0;i < len;++i)
    {
        int id = str[i] - 'a';
        if (p->nx[id] == NULL) return 0;
        p = p->nx[id];
    }
    return p->num;
}
void Delete(Trie *t)
{
    for (int i = 0;i < N;++i)
    {
        if (t->nx[i]!=NULL) Delete(t->nx[i]);
    }
    delete t;
}
int main()
{
     root = new Trie;
    init(root);
    int T,n,k;
    scanf("%d",&T);
    getchar();
    int i,j;
    char s[21];
    for(i=0;i<T;i++)
    {
     scanf("%s",s);
    for(j=0;j<strlen(s);j++)
        {
          Insert(s+j,i);
        }
    }
    scanf("%d",&n);
    getchar();
    for(i=0;i<n;i++)
    {
       scanf("%s",s);
       printf("%d\n",Find(s));
    }
    Delete(root);
    return 0;
}