Hat’s Words (字典树 + 智能指针shared_ptr)

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 298 Accepted Submission(s): 138 

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 <stdio.h> #include <iostream> #include <algorithm> #include <cmath> #include <vector> #include <string.h> #include <memory> using namespace std; string sss[55]; struct Node {    int flag = 0;    shared_ptr<Node> next[50]; }; void insert(const shared_ptr<Node>& root, const string& str) {    shared_ptr<Node> cur(root);    int i = 0;    while (i < str.size()) {        int k = str[i] - 'a';        if (cur->next[k] == NULL) {        Node *p = new Node;        cur->next[k].reset(p);        cur = cur->next[k];        } else        cur = cur->next[k];        i++;    }    cur->flag = 1;} bool search(const shared_ptr<Node>& root,const string& str) {    shared_ptr<Node> cur(root);    for (int i = 0; i < str.size(); ++i) {        if (cur->next[str[i] - 'a'] == NULL)            return false;        cur = cur->next[str[i] - 'a'];    }    return cur->flag; }  int main() {    shared_ptr<Node> root(new Node);    int cnt = 0;    while (cin >> sss[cnt]) {        insert(root, sss[cnt++]);    }    for (int i = 0; i < cnt; ++i) {    int len = sss[i].size();    for (int j = 1; j < len - 1; ++j) {        string str1(sss[i], 0, j);        string str2(sss[i],j);        if (search(root, str1) && search(root, str2)){            cout << sss[i] << endl;            break;            }        }    } }