Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

题意：给定一些单词（按字典序给出）， 按字典序输出所有满足条件的单词（条件为该单词由其它两个单词构成）

思路：先构造一颗字典树，，然后再依次判断该单词是否有其他两个单词组成，，判断方法为在该单词中找其中存在p->is为true的点，然后将点+1放入栈中，看是否这个点往后能组成一个其他单词，是则输出，，不是则不输出。。

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
using namespace std;

const int MAX = 50005;
char word[MAX][30];

struct node
{
	bool is;
	struct node *next[26];
	node()
	{
		is = false;
		memset(next, 0, sizeof(next));
	}
};

int insert(node *root, char *s)
{
	int i = 0;
	node *p = root;
	while(s[i])
	{
		if(p->next[s[i]-'a'] == NULL)
			p->next[s[i]-'a'] = new node;
		p = p->next[s[i]-'a'];
		i++;
	}
	p->is = true;
}

int search(node *root, char* s)
{
	node *p = root;
	int i = 0;
	stack<int> t;
	while(s[i])
	{
		if(p->next[s[i]-'a'] == NULL) return 0;
		p = p->next[s[i]-'a'];
		if(p->is && s[i+1])
			t.push(i+1);
		i++;
	}
	while(!t.empty())
	{
		bool ok = 1;
		i = t.top(); t.pop();
		p = root;
		while(s[i])
		{
			if(p->next[s[i]-'a'] == NULL)
			{
				ok = 0;
				break;
			}
			p = p->next[s[i]-'a'];
			i++;
		}
		if(ok && p->is) return 1;
	}
	return 0;
}

int main()
{
	int num = 0;
	node* root = new node();
	while(scanf("%s", word[num]) != EOF)
	{
		insert(root, word[num]);
		num++;
	}
	for(int i=0; i<num; i++)
		if(search(root, word[i]))
			printf("%s\n", word[i]);
	return 0;
}