HDU2846-Repository

最新推荐文章于 2018-01-22 20:02:42 发布

原创最新推荐文章于 2018-01-22 20:02:42 发布 · 248 阅读

0 ·

CC 4.0 BY-SA版权

HDU 同时被 2 个专栏收录

490 篇文章

订阅专栏

----字典树

16 篇文章

订阅专栏

本文介绍了一个字符串搜索模拟问题，该问题要求实现一个搜索系统来查询商品名称中包含特定子字符串的商品数量。通过构建字典树的数据结构，文章提供了一种高效的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Repository

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 4637 Accepted Submission(s): 1629

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

题意：给P个单词，然后有Q个询问，每个问给一个单词t，然后问在P个单词中，t是其字串的共有多少个。

解题思路：对于每个单词，把他的所有字串插入字符串，并在相应的位置+1. 注意一个单词如果有多个相同的字串，那这个字串也只能加一次。所有节点要多一个变量用来标记是否已经加过。

<pre name="code" class="cpp">#include <cstdio>  
#include <iostream>  
#include <algorithm>  
#include <cstring>  
#include <cmath>  
  
using namespace std;  
  
const int maxn=500005;  
int t,n,sum;  
char ch[100];  
  
struct tire  
{  
    int num[26];  
    int q,sum;  
    int operator[](int &x)  
    {  
        return num[x];  
    }  
    void sert(int x,int y)  
    {  
        num[x]=y;  
    }  
    void clea()  
    {  
        q=0;  
        sum=0;  
        memset(num,0,sizeof(num));  
    };  
} f[maxn];  
  
int main()  
{  
    while(~scanf("%d",&n))  
    {
        f[0].clea();
        for(int i=sum=0; i<n; i++)  
        {  
            scanf("%s",ch);  
            int len=strlen(ch);  
            for (int j=0; j<len; j++)  
            {  
                for (int k=j,l=0; k<len; k++)  
                {  
                    int x=ch[k]-'a';  
                    if(!f[l][x])  
                    {  
                        f[l].sert(x,++sum);  
                        f[sum].clea();  
                    }  
                    l=f[l][x];  
                    if(f[l].q!=i+1) f[l].sum++;  
                    f[l].q=i+1;  
                }  
            }  
        }  
        scanf("%d",&t);  
        while(t--)  
        {  
            scanf("%s",ch);  
            int j=0,len=strlen(ch);  
            for(int i=0; i<len; i++)  
            {  
                int x=ch[i]-'a';  
                j=f[j][x];  
                if(!j) break;  
            }  
            printf("%d\n",f[j].sum);  
        }  
    }  
    return 0;  
}