Repository（字典树，避免重复计数）

Myth丶恋晨 2022-08-01 12:22 119阅读 0赞

# Repository #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)  
Total Submission(s): 2865    Accepted Submission(s): 1096**

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][]

#include<cstdio>
    #include<iostream>
    #include<cstring>
    using namespace std;
    struct Tree
    {
        int num;
        int flag;
        Tree *next[26];
        Tree()
        {
            num=0;
            for(int i=0;i<26;i++)
            {
                next[i]=NULL;
            }
        }
    }*root;//建树！
    void insert(Tree *p,char *s,int n)
    {
        int i=0;
        while(s[i])
        {
            int x=s[i]-'a';
            if(p->next[x]==NULL)
            {
                p->next[x]=new Tree();
            }//如果没有就新建一个！
            p=p->next[x];
            if(p->flag!=n)//看他的记录如果不是n
            {
                p->flag=n;//标记新的flag
                p->num++;//数字加一！
            }//重复就算一个！
            i++;
        }
    }
    int find(Tree *p,char *s)
    {
        int ans,i=0;
        while(s[i])
        {
            int n=s[i]-'a';
            if(p->next[n]==NULL)
            {
                return 0;
            }//不重复就0
            else
            {
                p=p->next[n];
                ans=p->num;
            }
            i++;
        }
        return ans;
    }
    int main()
    {
        int n,m,stl,i,j;
        char s[30];
        root=new Tree();
        cin>>n;
        for(i=1;i<=n;i++)
        {
           cin>>s;
            stl=strlen(s);
            for(j=0;j<stl;j++)
            {
                insert(root,s+j,i);
            }
        }
     //   cout<<"   ok"<<endl;
        cin>>m;
        while(m--)
        {
           cin>>s;
            cout<<find(root,s)<<endl;
        }
    }

[2009 Multi-University Training Contest 4 - Host by HDU]: http://acm.hdu.edu.cn/search.php?field=problem&key=2009+Multi-University+Training+Contest+4+-+Host+by+HDU&source=1&searchmode=source