HDU2846 Repository 字典树

题目链接：HDU2846

Repository

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3868    Accepted Submission(s): 1401

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
   
   
   
   
   
   
   
   


   
   
   
   

题意：给一组字符串和一组模式串，问模式串在前面字符串中共出现多少次。

题目分析：字典树就可以了，不同的是对于同一个字符串进行插入操作时每次都向前进一位到末尾组成新串再进行插入。注意注意如aaa这种字符串在插入计数的时候可能会算重复，所以对每一个插入的字符串需要标记一下。

还有就是提交代码最好要C++，我之前提交好几次G＋＋都MLE了，换成C++立刻AC

//
//  main.cpp
//  HDU2846
//
//  Created by teddywang on 16/3/29.
//  Copyright © 2016年 teddywang. All rights reserved.
//

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef struct node{
    int flag;
    int num;
    node *next[26];
}trienode;
trienode *root;
char s[26];
int num,num2;

void insert_node(char *s,int num)
{
    trienode *r,*t=root;
    int len=strlen(s);
    for(int i=0;i<len;i++)
    {
        int buf=s[i]-'a';
        if(t->next[buf]==NULL)
        {
            r=new node;
            for(int j=0;j<26;j++)
            {
                r->next[j]=NULL;
            }
            r->flag=num;
            r->num=1;
            t->next[buf]=r;
            t=t->next[buf];
        }
        else
        {
            t=t->next[buf];
            if(t->flag!=num)
            {
                t->num=t->num+1;
                t->flag=num;
            }
        }
    }
}

int find(char *s)
{
    trienode *t=root;
    int len=strlen(s);
    for(int i=0;i<len;i++)
    {
        int buf=s[i]-'a';
        if(t->next[buf]==NULL)
            return 0;
        else t=t->next[buf];
    }
    return t->num;
}

void del(trienode *root)
{
    for(int i=0;i<26;i++)
    {
        if(root->next[i]!=NULL)
            del(root->next[i]);
    }
    free(root);
}

int main()
{
    root=new node;
    root->flag=root->num=0;
    for(int i=0;i<26;i++)
    {
        root->next[i]=NULL;
    }
    cin>>num;
    for(int i=0;i<num;i++)
    {
        memset(s,0,sizeof(s));
        scanf("%s",s);
        int len=strlen(s);
        char t[26];
        strcpy(t,s);
        for(int j=0;j<len;j++)
        {
            insert_node(t+j,i);
        }
    }
    cin>>num2;
    for(int i=0;i<num2;i++)
    {
        char s2[26];
        scanf("%s",s2);
        cout<<find(s2)<<endl;
    }
    del(root);
}