hdu 1247 Hat’s Words

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

思路：将每个单词进行拆分为两部分，看一下是否能在字典树里找到他。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAXNUM 26
using namespace std;
char words[50005][100];

typedef struct Trie
{
bool flag;
Trie *next[MAXNUM];
}Trie;

Trie *root;

void init()
{
root = (Trie *)malloc(sizeof(Trie));
root->flag = false;
for (int i = 0;i<MAXNUM;i++)
root->next[i] = NULL;
}

void insert(char *word)
{
Trie *tem = root;
while (*word != '\0')
{
if (tem->next[*word - 'a'] == NULL)
{
Trie *cur = (Trie *)malloc(sizeof(Trie));
for (int i = 0;i<MAXNUM;i++)
cur->next[i] = NULL;
cur->flag = false;
tem->next[*word - 'a'] = cur;
}
tem = tem->next[*word - 'a'];
word++;
}
tem->flag = true;
}

bool search(char *word)
{
Trie *tem = root;
for (int i = 0;word[i] != '\0';i++)
{
if (tem == NULL || tem->next[word[i] - 'a'] == NULL)
return false;
tem = tem->next[word[i] - 'a'];
}
return tem->flag;
}

int main()
{
init();
int t = 0;
char w[100];
while (scanf("%s", words[t]) != EOF)
{

insert(words[t]);
t++;
}
for (int i = 0;i<t;i++)
{
memset(w, '\0', sizeof(w));
for (int j = 0;words[i][j] != '\0';j++)
{
*(w + j) = *(words[i] + j);
if (search(w) && search((words[i] + j + 1)))
{
printf("%s\n", words[i]);
break;
}
}

}
return 0;
}