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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): 13752 Accepted Submission(s): 4919
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
Author
戴帽子的
解析：字典树。
#include <cstdio>#include <cstring>char s[50005][50];struct Node{ Node* pt[26]; bool isEnd;};Node* root;Node memory[100000];int allo;void trie_init(){ memset(memory, 0, sizeof(memory)); allo = 0; root = &memory[allo++];}void trie_insert(char str[]){ Node *p = root; for(int i = 0; str[i] != ‘\0‘; ++i){ int id = str[i]-‘a‘; if(p->pt[id] == NULL){ p->pt[id] = &memory[allo++]; } p = p->pt[id]; } p->isEnd = true;}bool trie_find(char str[]){ Node *p = root; for(int i = 0; str[i] != ‘\0‘; ++i){ int id = str[i]-‘a‘; if(p->pt[id] == NULL){ return false; } p = p->pt[id]; } return p->isEnd;}void getans(char str[]){ Node* p = root; for(int i = 0; str[i] != ‘\0‘; ++i){ int id = str[i]-‘a‘; if(p->pt[id] == NULL){ return; } else{ if(p->pt[id]->isEnd && str[i+1] != ‘\0‘){ bool ok = trie_find(str+i+1); if(ok){ printf("%s\n", str); return; } } } p = p->pt[id]; }}void solve(int n){ trie_init(); for(int i = 0; i < n; ++i){ trie_insert(s[i]); } for(int i = 0; i < n; ++i){ getans(s[i]); }}int main(){ int n = 0; while(gets(s[n])){ ++n; } solve(n); return 0;}
HDU 1247 Hat’s Words
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