【Trie】hdu 1247 Hat’s Words

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本文介绍如何使用Trie树数据结构解决题目，通过遍历字典中的单词，判断是否由两个已知单词拼接而成的'帽词'。代码演示了如何构建Trie树并查找这些特殊单词。

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22924 Accepted Submission(s): 8049

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

建了Trie树之后，直接枚举查找即可

代码

#include<bits/stdc++.h>
using namespace std;
char c[50005][30];
int ch[50005][30],k,tot,cnt[50005];
void insert(char *s)
{
	int now=0,len=strlen(s);
	for(int i=0;i<len;i++)
	{
		int dig=s[i]-'a';
		if(!ch[now][dig]) ch[now][dig]=++tot;
		now=ch[now][dig];
	}
	cnt[now]=1;
}
int check(int k,int pos)
{
	int now=0,len=strlen(c[k]);
	for(int i=pos;i<len;i++)
	{
		int dig=c[k][i]-'a';
		if(!ch[now][dig]) return 0;
		now=ch[now][dig]; 
	}
	if(cnt[now]>0) return 1;
	else return 0;
}
int find(int k)
{
	int now=0,len=strlen(c[k]);
	for(int i=0;i<len;i++)
	{
		int dig=c[k][i]-'a';
		if(cnt[now]>0 && check(k,i)) return 1;
		now=ch[now][dig]; 
	}
	return 0;
}
int main()
{
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	while(scanf("%s",c[++k])!=EOF)
	{
		insert(c[k]);
	}
	for(int i=1;i<=k;i++)
		if(find(i))
		{
			for(int j=0;j<strlen(c[i]);j++)
				printf("%c",c[i][j]);
			printf("\n");
		}
	return 0;
}