HDU 2846 - Repository

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

20adaeafagahaiajakaladsaddadeadfadgadhadiadjadkadlaes5badads

Sample Output

02011112

 

题意：先给出一个数字 N，后面 N 行给出 N 个字符串，后面一行给出一个数字 M，后面 M 行字符串，求出每个字符串在前面出现的次数。

是一个字符串的匹配问题，用字典树非常方便，上网看了下模板很快就过了。

#include <cstdio>#include <cstring>#include <iostream>using namespace std;struct tree{    int id;    int num;    tree *next[26];    tree ()    {        num = 0;        id = -1;        for (int i = 0; i < 26; ++i)            next[i] = NULL;    }};char s[25];tree *root = new tree;void Insert(char str[],int k){    int len=strlen(str);    tree* cur=root;    tree* q;    for(int i = 0; i < len ; i++)    {        int x = str[i]-'a';        if(cur->next[x] == NULL)        {            q = new tree;            q->id = k;            q->num = 1;            for(int j = 0; j < 26; j++)                q->next[j] = NULL;            cur->next[x] = q;        }        cur = cur->next[x];        if(cur->id != k)        {            cur->id = k;            cur->num++;        }    }}int Search(char str[]){    int len = strlen(str);    tree *p = root;    for(int i = 0; i < len; i++)    {        int x = str[i]-'a';        if(p->next[x])            p = p->next[x];        else            return 0;    }    return p->num;}int main(){    int n;    for(int i = 0; i < 26; i++)        root->next[0] = NULL;    root->num = 0;    root->id = -1;    scanf("%d", &n);    for(int i = 0; i < n; i++)    {        scanf("%s", s);        int len = strlen(s);        for(int j = 0; j < len; j++)        {            Insert(s + j, i);        }    }    int m;    scanf("%d", &m);    while(m--)    {        scanf("%s",s);        printf("%d\n", Search(s));    }    return 0;}