Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131070/65535 K (Java/Others)
Total Submission(s): 46494 Accepted Submission(s): 16531

Problem Description

Ignatius最近遇到一個難題,老師交給他很多單詞(只有小寫字母組成,不會有重複的單詞出現),現在老師要他統計出以某個字串為字首的單詞數量(單詞本身也是自己的字首).

Input

輸入資料的第一部分是一張單詞表,每行一個單詞,單詞的長度不超過10,它們代表的是老師交給Ignatius統計的單詞,一個空行代表單詞表的結束.第二部分是一連串的提問,每行一個提問,每個提問都是一個字串.

注意:本題只有一組測試資料,處理到檔案結束.

Output

對於每個提問,給出以該字串為字首的單詞的數量.

Sample Input

banana
band
bee
absolute
acm

ba
b
band
abc

Sample Output

2
3
1
0

Author

Ignatius.L

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字典樹的基礎應用。

#include<iostream>
 

#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long int
struct Trie {
    int v;
    Trie *next[26];
};
Trie root;
void createTrie(char *str)
{
    int len = strlen(str);
    Trie *p = &root, *q;
    for (int i = 0; i < len; i++)
    {
        int
 
 id = str[i] - 'a';
        if (p->next[id] == NULL)
        {
            q = (Trie *)malloc(sizeof(root));
            q->v = 1;
            for (int j = 0; j < 26; j++)
            {
                q->next[j] = NULL;
            }
            p->next[id] = q;
            p = p->next[id];
        }
        else
        {
            p->next[id]->v++;
            p = p->next[id];
        }
    }
}
int findTrie(char *str)
{
    int len = strlen(str);
    Trie *p = &root;
    for (int i = 0; i < len; i++)
    {
        int id = str[i] - 'a';
        p = p->next[id];
        if (p == NULL)
            return 0;//chaxunwuguo
    }
    return p->v;
}
int main()
{
    char str[15];
    int i;
    for (i = 0; i < 26; i++)
    {
        root.next[i] = NULL;
    }
    while (gets(str)&&str[0]!='\0')
    {
        createTrie(str);
    }
    memset(str, 0, sizeof str);
    while (~scanf("%s", str))
    {
        int ans = findTrie(str);
        printf("%d\n", ans);
    }
    return 0;
}