hdu1247Hat’s Words

最新推荐文章于 2021-08-05 13:35:01 发布

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#struct #concatenation #dictionary #insert #output #input

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本文介绍了一种使用字典树（Trie）来找出所有由字典中两个单词拼接而成的特殊单词（Hat's Words）。通过构建字典树并遍历每个单词来检查其是否能被拆分为两个存在于字典中的子单词。

字典树，感觉有点暴力。

Hat’s Words

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 4 Accepted Submission(s) : 4

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Problem Description

A hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.
You are to find all the hat’s words in a dictionary.

Input

Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 50,000 words.
Only one case.

Output

Your output should contain all the hat’s words, one per line, in alphabetical order.

Sample Input

a
ahat
hat
hatword
hziee
word

Sample Output

ahat
hatword
<code>

#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
using namespace std;
char str[50005][100],str1[100],str2[100];
struct dirtree 
{
   struct dirtree *child[26];
   bool hash;
};
struct dirtree *root;
void Insert(char*source)
{
     int i,j,len;
     len=strlen(source);
     if(len==0)return ;
     struct dirtree *current,*newnode;
     current=root;
     for(i=0;i<len;i++)
     {
        if(current->child[source[i]-'a']!=0)
        {
           current=current->child[source[i]-'a'];
           if(i==len-1)
           {
              current->hash=true;
              break;
           }
           
        }
        else
        {
           newnode=(struct dirtree*)malloc(sizeof(struct dirtree));
           for(j=0;j<26;j++)
           {
              newnode->child[j]=0;
              newnode->hash=false;
           }
           current->child[source[i]-'a']=newnode;
           current=newnode;
           if(i==len-1)
           {
              current->hash=true;
              break;
           }
            
        }
     }
}
bool Find(char *source)
{
    struct dirtree *current;
    int len=strlen(source);
    if(len==0)return 0;
    current=root;
    for(int i=0;i<len;i++)
    {
       if(current->child[source[i]-'a']!=0)
          current=current->child[source[i]-'a'];
       else
          return false;
    }
    return current->hash;
}
int main()
{
 
   int len = 0;
   int i, j, k, len1;
   root = (struct dirtree *)malloc(sizeof(struct dirtree));
   for(i = 0; i < 26; i++)
   {
      root->child[i] = 0;
      root->hash = false;
   }
   while(scanf("%s",str[len])!=EOF)
   {
       Insert(str[len]);
       len++;
   }
   for(i=0;i<len;i++)
   {
     len1=strlen(str[i]);
     for(j=1;j<len1;j++)
     {
        for(k=0;k<j;k++)
        {
           str1[k]=str[i][k];
        }
        str1[k]='/0';
        int t=0;
        for(k=j;k<len1;k++)
        {
           str2[t++]=str[i][k];
        }
        str2[t]='/0';
        if(Find(str1)&&Find(str2))
        {
           printf("%s/n",str[i]);
           break;
        }
     }
   } 
   //system("pause");
   return 0;
   
}