hdu2846

Repository

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2136    Accepted Submission(s): 791

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

 

思路：把每个单词X1X2X3……分别以X1，X2，X3为开头存入字典树中，然后要进行标记，同一个单词不能多次计数

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<malloc.h>
#define MAX 26
using namespace std;
struct tree
{
    struct tree *child[26];
    int v;//记录包含该结点的单词个数
    int flag; //最后一次经过此结点的商品ID
}*root;
void init()
{
    int i;
    root=(struct tree*)malloc(sizeof(struct tree));
    for(i=0; i<26; i++)
    {
        root->child[i]=0;
        root->v=0;
        root->flag=-1;
    }
    return ;
}
void build(char *str,int d)
{
    int id,i,j,len;
    len=strlen(str);
    struct tree *p=root,*q;
    for(i=0; i<len; i++)
    {
        id=str[i]-'a';
        if(p->child[id]==0)
        {
            q=(struct tree*)malloc(sizeof(struct tree));
            for(j=0; j<26; j++)
                q->child[j]=0;
            p->child[id]=q;
            p=p->child[id];
            p->v=0;
            p->flag=-1;
        }
        else
            p=p->child[id];
        if(p->flag!=d)//如果当前结点的商品编号不等于要插入商品的编号，则计数器v++，并且重新置编号的值
        {
            p->flag=d;
            p->v++;
        }
    }
}
int find(char *str)
{
    int len=strlen(str);
    int i,j,id;
    struct tree *p=root,*q;
    for(i=0; i<len; i++)
    {
        id=str[i]-'a';
        if(p->child[id]==0)
            return 0;
        p=p->child[id];
    }
    return p->v;
}
int main()
{
    int i,j,len,tt,t;
    char a[105],b[105];
    scanf("%d",&tt);
    init();
    for(i=0;i<tt;i++)
    {
        scanf("%s",b);
        len=strlen(b);
        for(j=0; j<len; j++)
            build(b+j,i);//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tree树中
    }
    scanf("%d",&t);
    for(i=0; i<t; i++)
    {
        scanf("%s",a);
        printf("%d ",find(a));
    }
    return 0;
}