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): 2205    Accepted Submission(s): 804

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：给一些单词（以字典序输入），找出那些可以分成另外的两个单词的单词，以字典序输出；

源代码：（15MS）

#include<iostream>
#include<string.h>
using namespace std;

const int NODEMAX=26;
const int MAX=50005;
char word[MAX][16];  //这道题16就足够了

struct TrieNode
{
   bool is;
   TrieNode *next[NODEMAX];
   TrieNode()
   {
      is=false;
      memset(next,0,sizeof(next));
   }
};

void InsertTrie(TrieNode *pRoot,char s[])
{
    int i;
    TrieNode *p=pRoot;
    i=0;
    while(s[i])
    {
         int k=s[i]-'a';
         if(p->next[k]==NULL)
            p->next[k]=new TrieNode();
         i++;
         p=p->next[k];
    }
    p->is=true; //该结点是单词的尾
}

bool Search(TrieNode *pRoot,char s[])
{
	int i,top=0,stack[1000];
	TrieNode *p=pRoot;
	i=0;
	while(s[i])
	{
		int k=s[i++]-'a';
		if(p->next[k]==NULL)
			return 0;
		p=p->next[k];
		if(p->is && s[i]) //找到该单词含有子单词的分隔点
			stack[top++]=i;//入栈
	}
	while(top)//从可能的分割点去找
	{
	    bool ok=1;
		i=stack[--top];
		p=pRoot;
		while(s[i])
		{
			int k=s[i++]-'a';
			if(p->next[k]==NULL)
			{
				ok=false;
				break;
			}
			p=p->next[k];
		}
		if(ok && p->is)//找到最后,并且是单词的
			return 1;
	}
	return 0;
}

int main()
{
    int i=0;
    TrieNode *pRoot=new TrieNode();
    while(gets(word[i]))
    {
        InsertTrie(pRoot,word[i]);
        i++;
	}
	for(int j=0;j<i;j++)
		if(Search(pRoot,word[j]))
			cout<<word[j]<<endl;
    return 0;
}