Hat's words (Dictionary tree)

Problem descriptiona 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.
 
Inputstandard 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.
 
Outputyour output shoshould contain all the hat's words, one per line, in alphabetical order.
Sample Input

aahathathatwordhzieeword

 
Sample output

ahathatword

 

The question is to give you some words. You need to find the words that can be composed of the other two words. There are more than 50000 pieces of data, and the dictionary tree

# Include <iostream> # include <cstring> # include <cstdio> # include <algorithm> using namespace STD; typedef struct trie {int flag; trie * Next [26];} trie; trie * root; int flag; char a [50000] [50]; void Inti () // initialization {root = (trie *) malloc (sizeof (trie )); root-> flag = 0; For (INT I = 0; I <26; I ++) Root-> next [I] = NULL;} void Charu (char *) // create a dictionary tree to insert the word {trie * P = root, * q; int L = strlen (a); For (INT I = 0; I <L; I ++) {int id = A [I]-'A'; If (p-> next [ID] = NULL) {q = (trie *) malloc (sizeof (trie); q-> flag = 0; For (INT I = 0; I <26; I ++) q-> next [I] = NULL; p-> next [ID] = Q;}/* else if (p-> next [ID]-> flag) {flag = 1 ;} */P = p-> next [ID];} p-> flag = 1;} int CZ (char * A) // find the word {trie * P = root; int Len = strlen (a); For (INT I = 0; I <Len; I ++) {int id = A [I]-'A '; P = p-> next [ID]; If (P = NULL) return 0;} return p-> Fla G;} int SF (trie * P) // you can skip this topic {// trie * P = root; If (P = NULL) return 0; for (INT I = 0; I <26; I ++) {If (p-> next [I]! = NULL) SF (p-> next [I]);} Free (p); Return 0;} int main () {Inti (); int n = 0; while (~ Scanf ("% s", & A [n]) {// flag = 0; // printf ("% s", a [I]); charu (A [n]); N ++;} For (INT I = 0; I <n; I ++) {int Len = strlen (A [I]); for (Int J = 0; j <Len; j ++) {char str1 [50] = {'\ 0 '}, str2 [50] = {'\ 0'}; // create two temporary arrays strncpy (str1, a [I], J); // split the word I, put strncpy (str2, a [I] + J, len-j) in two temporary arrays respectively; If (CZ (str1) & CZ (str2 )) // search through the search function. If both returns are 1, {printf ("% s \ n", a [I]); break;} can be formed ;}}} SF (Root); // release the memory (this question can be left blank) return 0 ;}