제출 #1171014

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1171014		ortsac	동굴 (IOI13_cave)	C++20	100 / 100	478 ms	628 KiB

cave

#include <bits/stdc++.h>
#include "cave.h"

using namespace std;

int n;
vector<int> found;
vector<int> state;
vector<int> ans;

void exploreCave(int N) {
    n = N;
    state.resize(n);
    ans.resize(n);
    found.resize(n);
    for (int i = 0; i < n; i++) {
        // we gonna find the switch that is connected to i, and the state of that switch
        // lets start be finding the state
        int s[n];
        for (int j = 0; j < n; j++) {
            if (found[j]) s[j] = state[j]; // j is the answer to some previous door
            else s[j] = 0;
        }
        int x = tryCombination(s);
        int istate = 1;
        if ((x == -1) || (x > i)) istate = 0; // i is open with 0
        // now we have to find the switch that is connected to i
        // binary search: having a group of all switches that could be connected to i
        // test half with istate, the other half with !istate
        // if the door is open, it is on the first half
        // else it is on the second half
        // detail: make sure that all the doors before i are already open
        vector<int> pos;
        for (int j = 0; j < n; j++) if (!found[j]) pos.push_back(j);
        while (pos.size() > 1) {
            // ok, now just do this
            for (int j = 0; j < n; j++) {
                if (found[j]) s[j] = state[j]; // j is the answer to some previous door
                else s[j] = 0;
            }
            vector<int> h1, h2;
            int t = pos.size();
            for (int j = 0; j < (t/2); j++) {
                h1.push_back(pos[j]);
                s[pos[j]] = istate;
            }
            for (int j = (t/2); j < t; j++) {
                h2.push_back(pos[j]);
                s[pos[j]] = (1 - istate); // opposite of istate
            }
            x = tryCombination(s);
            if ((x == -1) || (x > i)) pos = h1;
            else pos = h2;
        }
        found[pos.back()] = 1;
        state[pos.back()] = istate;
        ans[pos.back()] = i;
    }
    int s[n];
    int d[n];
    for (int i = 0; i < n; i++) s[i] = state[i];
    for (int i = 0; i < n; i++) d[i] = ans[i];
    answer(s, d);
}

• Subtask 1: 12 / 12
• Subtask 2: 13 / 13
• Subtask 3: 21 / 21
• Subtask 4: 30 / 30
• Subtask 5: 24 / 24