hdu1251(字典树)

统计难题

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131070/65535 K (Java/Others)
Total Submission(s): 15521    Accepted Submission(s): 6640

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

 

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
using namespace std;
struct dictree
{
    struct dictree *child[26];
    int n;
};
struct dictree *root;
void insert(char *source)
{
    int i,j;
    int len,id;
    struct dictree *current,*newnode;
    len=strlen(source);
    current=root;
    for(i=0; i<len; i++)
    {
        id=source[i]-'a';
        if(current->child[id]!=0)
        {
            current=current->child[id];
            current->n++;
        }
        else
        {
            newnode=(struct dictree *)malloc(sizeof(struct dictree));
            for(j=0; j<26; j++)
                newnode->child[j]=0;
            current->child[id]=newnode;
            current=newnode;
            current->n=1;
        }
    }
}
int find(char *source)
{
    int i,id;
    int len;
    struct dictree *current;
    len=strlen(source);
    if(len==0)
        return 0;
    current=root;
    for(i=0; i<len; i++)
    {
        id=source[i]-'a';
        if(current->child[id]!=0)
            current=current->child[id];
        else
            return 0;
    }
    return current->n;
}
int main()
{
    char temp[11];
    int i;
    root=(struct dictree *)malloc(sizeof(struct dictree));
    for(i=0; i<26; i++)
    root->child[i]=0;
    root->n=0;
    while(gets(temp),strcmp(temp,"")!=0)
        insert(temp);
    while(gets(temp)!=0)
        printf("%d\n",find(temp));
    return 0;

}