Submission #159559

#	Time	Username	Problem	Language	Result	Execution time	Memory
159559		rama_pang	Unscrambling a Messy Bug (IOI16_messy)	C++14	100 / 100	5 ms	636 KiB

messy

#include "messy.h"
#include <bits/stdc++.h>
using namespace std;

/*  Idea: Use Divide and Conquer, set each bits in interval le...ri to 1 once, adding it to data structure.
    For all bits outside interval, set it to 1.
    Then when checking, you can see which elemnts belong to interval le...mid and (mid + 1)...ri, then
    apply DnC normally.
*/

void add_permutation(int le, int ri, int N) {
    string element;
    for (int i = 0; i < N; i++) element.push_back('1');
    for (int i = le; i <= ri; i++) element[i - 1] = '0';
    if (le >= ri) return;

    int mid = (le + ri) / 2;
    for (int i = le; i <= mid; i++) {
        element[i - 1] = '1';
        add_element(element);
        element[i - 1] = '0';
    }

    add_permutation(le, mid, N);
    add_permutation(mid + 1, ri, N);

}

void get_permutation(int le, int ri, vector<int> forbidden, vector<int> &ans, int N) {
    string element;
    for (int i = 0; i < N; i++) element.push_back('0');
    for (auto i : forbidden) element[i - 1] = '1';

    if (le == ri) {
        for (int i = 0; i < N; i++) {
            if (element[i] == '0') {
                ans[i] = le - 1;
                return;
            }
        }
    }

    vector<int> left_forbidden = forbidden, right_forbidden = forbidden;
    for (int i = 1; i <= N; i++) {
        if (element[i - 1] == '1') continue;
        element[i - 1] = '1';
        if (check_element(element)) {
            right_forbidden.push_back(i);
        } else {
            left_forbidden.push_back(i);
        }
        element[i - 1] = '0';
    }

    int mid = (le + ri) / 2;
    get_permutation(le, mid, left_forbidden, ans, N);
    get_permutation(mid + 1, ri, right_forbidden, ans, N);

}

vector<int> restore_permutation(int n, int w, int r) {
    vector<int> res(n);
    add_permutation(1, n, n);
    compile_set();
    get_permutation(1, n, vector<int>(), res, n);
    
    return res;
}

• Subtask 1: 20 / 20
• Subtask 2: 18 / 18
• Subtask 3: 11 / 11
• Subtask 4: 21 / 21
• Subtask 5: 30 / 30