제출 #415988

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
415988		fvogel499	Type Printer (IOI08_printer)	C++14	80 / 100	1093 ms	125892 KiB

printer

/*
File created on 06/01/2021 at 20:53:10.
Link to problem: https://oj.uz/problem/view/IOI08_printer
Description: use a trie structures
Time complexity: O(N)
Space complexity: O(N)
Status: DEV
Copyright: Ⓒ 2021 Francois Vogel
*/

#include <iostream>
#include <cmath>
#include <vector>
#include <bitset>
#include <queue>
#include <cstring>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <algorithm>

using namespace std;

#define pii pair<int, int>
#define f first
// #define s second

#define pb push_back
#define ins insert
#define cls clear

#define ll long long

vector<char> steps;
int proc;

struct Node {
    Node(char lt, int ld) {
        t = lt;
        d = ld;
        md = d;
        words = 0;
        for (int i = 0; i < 26; i++) exist[i] = nullptr;
    }
    void add(string& s) {
        if (s.empty()) {
            words++;
            return;
        }
        int i = s[s.size()-1]-'a';
        if (exist[i] == nullptr) {
            exist[i] = new Node(char(i)+'a', d+1);
            c.pb(exist[i]);
        }
        s.pop_back();
        exist[i]->add(s);
        md = max(md, exist[i]->md);
    }
    int words, d, md;
    char t;
    Node* exist [26];
    vector<Node*> c;
};

void dfs(Node* i) {
    for (int j = 0; j < i->words; j++) {
        steps.pb('P');
        proc--;
    }
    for (Node* j : i->c) {
        if (j->md == i->md) continue;
        steps.pb(j->t);
        dfs(j);
        if (proc > 0) steps.pb('-');
    }
    for (Node* j : i->c) {
        if (j->md != i->md) continue;
        steps.pb(j->t);
        dfs(j);
        if (proc > 0) steps.pb('-');
    }
}

signed main() {
    cin.tie(0);
    // ios_base::sync_with_stdio(0);

    Node* root = new Node('$', 0);

    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        string ls;
        cin >> ls;
        reverse(ls.begin(), ls.end());
        root->add(ls);
    }

    proc = n;
    dfs(root);

    cout << steps.size() << endl;
    for (char i : steps) cout << i << endl;

    int d = 0;
    d++;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10