hdu 1247 (字典树入门)

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12149    Accepted Submission(s): 4338

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

题意：找出能有已知给定两个单词组成的单词。
思路：先插入每个单词进入字典树，然后分别对每个单词进行查找，find找到前缀后再利用find2找到后缀，当后缀对应某个单词时，则证明此单词由两个字典中单词组成
注意输入到文件尾

package 字典树;

import java.util.Scanner;

class Trie{
    private Node root;
    public Trie() {
        root = new Node();
    }
    public void insert(String s){
        Node t =root;
        for(int i=0;i<s.length();i++){
            if(t.nodes[s.charAt(i)-'a']==null){
                Node node = new Node();
                t.nodes[s.charAt(i)-'a'] = node;
            }
            t = t.nodes[s.charAt(i)-'a'];
        }
        t.isword = true;
    }
    public boolean find(String str){
        Node t = root;
        for(int i=0;i<str.length();i++){
            if(t.nodes[str.charAt(i)-'a']!=null){
                if(t.isword){
                    String str1 = str.substring(i, str.length());
                    //System.out.println(str1);
                    if(find2(str1)){
                        return true;
                    }
                }
                t = t.nodes[str.charAt(i)-'a'];
            }
            else return false;
        }
        return false;
    }
    private boolean find2(String str) {
        Node t = root;
        for(int i=0;i<str.length();i++){
            if(t.nodes[str.charAt(i)-'a']==null){
                return false;
            }
            t = t.nodes[str.charAt(i)-'a'];
        }
        if(t.isword) return true;
        return false;
    }
    public 
    class Node {
        Node [] nodes;
        boolean isword;  //标记末尾
        public Node() {
            isword = false;
            nodes = new Node[26];
        }
    }
}
public class hdu_1247 {
    public static void main(String[] args) {
        Scanner sc =new Scanner(System.in);
        Trie t = new Trie();
        String [] str  = new String [50005];
        int k=0;
         
        while(sc.hasNext())
        {
            str[k] = new String();
            str[k] = sc.next();
            t.insert(str[k++]);
        }
        for(int i=0;i<k;i++){
            if(t.find(str[i])){
                System.out.println(str[i]);
            }
        }
    }
}