hdu--1247--Hat’s Words（一般）

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5772    Accepted Submission(s): 2154

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

 

 

 

 

#include<iostream>

#include<cstring>

#include<cstdio>

using namespace std;

const int MAXN=26;

const int MAX=50005;

char word[MAX][27];



struct node

{

   bool is;

   node *next[MAXN];

   node()

   {

      is=false;

      memset(next,0,sizeof(next));

   }

};



void Insert(node *rt,char *s)

{

    int i=0;

    node *p=rt;

    while(s[i])

    {

         int k=s[i++]-'a';

         if(p->next[k]==NULL)

            p->next[k]=new node();

         p=p->next[k];

    }

    p->is=true; //该结点是单词的尾

}



bool Search(node *rt,char s[])

{

	int i=0,top=0,stack[1000];

	node *p=rt;

	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割点 

			stack[top++]=i;//入栈

	}

	while(top)//从可能的分割点去找

	{

	    bool ok=1;

		i=stack[--top];

		p=rt;

		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;

    node *rt=new node();

    while(gets(word[i]))

    {

        Insert(rt,word[i]);

        i++;

	}

	for(int j=0;j<i;j++)

		if(Search(rt,word[j]))

		   printf("%s\n",word[j]);

    return 0;

}