hdu 1247 Hat’s Words

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3431    Accepted Submission(s): 1303

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、速度够快的、就是太占空间呀

//判断每个单词是否由给的单词表里的2个单词组成

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#define N 60000
#define Size 26
using namespace std;
struct Tire
{
  bool isword;
  Tire *next[Size];
}node [N];
char s[50000][20];
int nu;
void insert(Tire *&root,char *word)
{
    Tire *p=root;
    while(*word)
    {
        if(p->next[*word-'a']==NULL)
            p->next[*word-'a']=&node[++nu];
         p=p->next[*word-'a'];
        word++;
    }
    p->isword=true;
}
bool ask(Tire *root,char *word)
{
    Tire *p=root;
    while(*word)
    {
        if(p->next[*word-'a']==NULL)
          return false;
        p=p->next[*word-'a'];
      word++;
    }
    if(p->isword) return true;
    return false;
}
bool query(Tire *root,char *word)
{
    Tire *p=root;
    while(*word)
    {
        if(p->next[*word-'a']==NULL)
          return false;
        p=p->next[*word-'a'];
        if(p->isword&&ask(root,word+1))
           return true;
      word++;
    }
    return false;
}
int main()
{
    int n=-1;
    Tire *root=&node[0];
    root->isword=false;
    while(scanf("%s",s[++n])!=EOF)
    {
       insert(root,s[n]);
    }
    for(int i=0;i<=n;i++)
     if(query(root,s[i]))
      printf("%s\n",s[i]);

    return 0;
}