Submission #1248049

#TimeUsernameProblemLanguageResultExecution timeMemory
1248049rythm_of_knightMinerals (JOI19_minerals)C++17
0 / 100
1 ms424 KiB
#include "minerals.h"
#include <bits/stdc++.h>
#define all(x) x.begin(), x.end()
using namespace std;

const int MXN = 43043, inf = 1e7;
vector <int> a, b;
static int cnt = 0, n, m;
int lg[MXN << 1], dp[MXN][2], mid[MXN][2];

void rec(int l, int r, int k, vector <int> q) {
    if (l == r)
        return Answer(a[l], q[0]), void();
    int m = l + mid[r - l + 1][k], lk = k, rk = k;
    vector <int> lft, rgt;
    int lneed = m - l + 1, rneed = r - m;
    if (k) {
        for (int i = m + 1; i <= r; i++)
            cnt = Query(a[i]);
        rk ^= 1;
    } else {
        for (int i = l; i <= m; i++)
            cnt = Query(a[i]);
        lk ^= 1;
    }
    for (int j : q) {
        if (lft.size() == lneed) {
            rgt.push_back(j);
        } else if (rgt.size() == rneed) {
            lft.push_back(j);
        } else {
            int tmp = Query(j);
            if (tmp == cnt) {
                lft.push_back(j);
            } else {
                rgt.push_back(j);
            }
            cnt = tmp;
        }
    }
    rec(l, m, lk, lft);
    rec(m + 1, r, rk, rgt);
}

void Solve(int N) {
    n = N, m = 2 * n;
    lg[0] = -1;
    for (int i = 1; i <= m; i++)
        lg[i] = lg[i >> 1] + 1;
    for (int i = 1; i <= m; i++) {
        int tmp = Query(i);
        if (tmp != cnt)
            a.push_back(i);
        else
            b.push_back(i);
        cnt = tmp;
    }
    for (int i = 2; i <= n; i++) {
        dp[i][0] = dp[i][1] = inf;
        int md = i >> 1;
        int tl = max(0, md - 3000);
        int tr = min(i - 2, md + 3000);
        for (int j = tl; j <= tr; j++) {
            if (dp[i][1] > dp[j + 1][1] + dp[i - j - 1][0] + i - j - 1) {
                dp[i][1] = dp[j + 1][1] + dp[i - j - 1][0] + i - j - 1;
                mid[i][1] = j;
            }
        }
        dp[i][1] += i - 1;
        for (int j = tl; j <= tr; j++) {
            if (dp[i][0] > dp[j + 1][1] + dp[i - j - 1][0] + j + 1) {
                dp[i][0] = dp[j + 1][1] + dp[i - j - 1][0] + j + 1;
                mid[i][0] = j;
            }
        }
        dp[i][0] += i - 1;
    }
    cout << dp[n][0] << ' ' << mid[n][0]  << '\n';
    rec(0, n - 1, 1, b);
}
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