HDU 1247 Hat’s Words（字典树）

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3565    Accepted Submission(s): 1355

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

 

 

Author

戴帽子的

 

 

Recommend

Ignatius.L

 

 

字典树的题目。

就是看单词是不是可以由其他两个单词合并起来。

建树。

然后将每个单词进行划分，看存不存在。

 

// HDU 1247 Hat¡¯s Words
//×ÖµäÊ÷

#include<iostream>
#include<string.h>
#include<stdlib.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
const int MAX=26;

typedef struct Trie_Node
{
    bool isWord;
    struct Trie_Node *next[MAX];
}Trie;
char s[50010][50];

void insert(Trie *root,char *word)
{
    Trie *p=root;
    int i=0;
    while(word[i]!='\0')
    {
        if(p->next[word[i]-'a']==NULL)
        {
           // Trie *temp=(Trie *)malloc(sizeof(Trie));
           Trie *temp=new Trie;
            for(int j=0;j<MAX;j++)
             temp->next[j]=NULL;
            temp->isWord=false;
            p->next[word[i]-'a']=temp;
        }
        p=p->next[word[i]-'a'];
        i++;
    }
    p->isWord=true;
}
bool search(Trie *root,char *word)//查找单词是否勋在
{
    Trie *p=root;
    for(int i=0;word[i]!='\0';i++)
    {
        if(p->next[word[i]-'a']==NULL)//?
          return false;
        p=p->next[word[i]-'a'];
    }
    return p->isWord;
}
void del(Trie *root)
{
    for(int i=0;i<MAX;i++)
    {
        if(root->next[i]!=NULL)
          del(root->next[i]);
    }
    free(root);
}
int main()
{
   // freopen("in.txt","r",stdin);
   // freopen("out.txt","w",stdout);
    int i,j;
    int cnt=0;
    char str[50];
  //  Trie *root=(Trie *)malloc(sizeof(Trie));
  Trie *root=new Trie;
    for(i=0;i<MAX;i++)
       root->next[i]=NULL;
    root->isWord=false;
    while(scanf("%s",s[cnt])!=EOF)
    {
        insert(root,s[cnt++]);
    }
    for(i=0;i<cnt;i++)
    {
        int len=strlen(s[i]);
        for(j=1;j<len-1;j++)
        {
            char temp1[50]={'\0'};
            char temp2[50]={'\0'};
            strncpy(temp1,s[i],j);
            strncpy(temp2,s[i]+j,len-j);
            if(search(root,temp1)&&search(root,temp2))
            {
                printf("%s\n",s[i]);
                break;
            }
        }
    }
    del(root);
    return 0;
}