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.
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.
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
题目意思是 给你一定的单词。要你找出可以通过其它两个单词组成的单词。数据有50000多。字典树了
#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()//初始化 { root = (Trie*)malloc(sizeof(Trie)); root->flag = 0; for (int i = 0; i < 26; i++) root->next[i] = NULL; } void ChaRu(char *a)//建立字典树 插入单词 { 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)//查找单词 { 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->flag; } int SF(Trie *p)//释放空间 这个题不用也能够过 { // 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]={'