HDU 1251 统计难题【字典树】

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最近遇到一个难题,老师交给他很多单词(只有小写字母组成,不会有重复的单词出现),现在老师要他统计出以某个字符串为前缀的单词数量(单词本身也是自己的前缀).

Input

输入数据的第一部分是一张单词表,每行一个单词,单词的长度不超过10,它们代表的是老师交给Ignatius统计的单词,一个空行代表单词表的结束.第二部分是一连串的提问,每行一个提问,每个提问都是一个字符串.

注意:本题只有一组测试数据,处理到文件结束.

Output

对于每个提问,给出以该字符串为前缀的单词的数量.

Sample Input

banana
band
bee
absolute
acm

ba
b
band
abc

Sample Output

2
3
1
0

Author

Ignatius.L

Recommend

Ignatius.L | We have carefully selected several similar problems for you: 1075 1247 1671 1298 1800

字典树的基础应用。

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>using namespace std;#define ll long long intstruct 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;}