Hat’s Words
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5201 Accepted Submission(s): 1940
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
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <cstring> #define MAXWORDS 26 char s[50000][50]; typedef struct node { bool isWord; node *next[MAXWORDS]; }TreeNode; void ZeroNode(TreeNode *&Node) { for (int i = 0; i < MAXWORDS; i++) { Node->next[i] = NULL; } } void Insert(TreeNode *&pRoot, char *pstr) { int nLen = strlen(pstr); TreeNode *p = pRoot; for (int i = 0; i < nLen; i++) { int index = pstr[i] - 'a'; if (p->next[index] == NULL) { TreeNode *Node = (TreeNode *)malloc(sizeof(TreeNode)); ZeroNode(Node); Node->isWord = false; p->next[index] = Node; } p = p->next[index]; } p->isWord = true; } bool Search(TreeNode *pRoot, char *pstr) { int nLen = strlen(pstr); for (int i = 0; i < nLen; i++) { int index = pstr[i] - 'a'; if (pRoot->next[index] != NULL) { pRoot = pRoot->next[index]; } else { return false; } } return pRoot->isWord; } void Delete(TreeNode *pRoot) { for (int i = 0; i < MAXWORDS; i++) { if (pRoot->next[i] != NULL) { Delete(pRoot->next[i]); } } free(pRoot); } int main() { int Count = 0; TreeNode *pRoot = (TreeNode *)malloc(sizeof(TreeNode)); ZeroNode(pRoot); while(scanf("%s", s[Count]) != EOF) { Insert(pRoot, s[Count++]); } for (int i = 0; i < Count; i++) { int nLen = strlen(s[i]); for (int j = 1; j < nLen; j++) { char temp1[50] = {'\0'}; char temp2[50] = {'\0'}; strncpy(temp1, s[i], j); strncpy(temp2, s[i] + j, nLen - j); if (Search(pRoot, temp1) && Search(pRoot, temp2)) { printf("%s\n", s[i]); break; } } } Delete(pRoot); return 0; }