HDU 1247 Hat’s Words(辞書ツリー)

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
帽子をかぶった
 
 
Recommend
Ignatius.L
 
 
辞書の木のテーマ.
単語が他の2つの単語から結合できるかどうかを見ることです.
木を建てる.
それから各単語を区分して、存在しないかどうかを見ます.
 

// 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)//        

{

    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
",s[i]);

                break;

            }

        }

    }

    del(root);

    return 0;

}