【Trie】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): 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;
}