HDU-1247 Hat’s Words （Trie 字典树）

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16368    Accepted Submission(s): 5835

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

aahathathatwordhzieeword

 

Sample Output

ahathatword

 

#include <bits/stdc++.h>using namespace std;int ch[500000][26], st[50002 * 26 * 20], tot; //st[i] == 1表示结点i是某个单词的结尾string s[50005];void insert(string s){int k = 0, y;for(int i = 0; i < s.size(); ++i){y = s[i] - 'a';if(ch[k][y] == 0){ch[k][y] = ++tot;}k = ch[k][y];}st[k] = 1;}bool check(string s, int pos){int k = 0, y;for(int i = pos; i < s.size(); ++i){y = s[i] - 'a';if(ch[k][y] == 0) return false;k = ch[k][y];}if(st[k]) return true;return false;}bool query(string s){int k = 0, y;for(int i = 0; i < s.size(); ++i){y = s[i] - 'a';if(ch[k][y] == 0){return false;}k = ch[k][y];if(st[k] == 1){if(check(s, i + 1)){return true;}}}return false;}int main(){int x = 1;tot = 0;while(cin>>s[x]){insert(s[x]);x++;}for(int i = 1; i < x; ++i){if(query(s[i])){cout<<s[i]<<endl;}}}/*题意：一共50000个单词，问哪些单词可以由其余单词中的两个拼接而成。比如hmc可以由hm和c拼接成。单词长度没说。。。思路：trie的小应用，对所有的单词建trie，然后在这棵trie上枚举所有单词，当遇到st值为1时再判断接下来的部分是否是一个完整的单词即可。这个方法很容易想到，应该可以被优化，欢迎指点。。。*/