Submission #1095206

#	Time	Username	Problem	Language	Result	Execution time	Memory
1095206		sunflower	Stove (JOI18_stove)	C++14	100 / 100	73 ms	51548 KiB

stove

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

#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)

template <class X, class Y>
    bool minimize(X &x, Y y) {
        if (x > y) return x = y, true;
        else return false;
    }

int n, k;

const int maxn = (int) 1e5 + 7;

int a[maxn + 2];

namespace subtask1 {
    bool check() {
        return (n <= 20);
    }

    const int inf = (int) 1e9 + 7;
    const int N = (int) 20 + 7;

    int dp[N + 2][N + 2]; // dp[i][j]: tong time khi xet i nguoi dau tien va da bat k lan;

    void solve() {
        FOR(i, 0, n)
            FOR(j, 0, k) dp[i][j] = inf;

        dp[0][0] = 0;
        FOR(turn, 1, k) {
            FOR(i, 1, n) {
                FOR(j, 1, i) {
                    // use: i -> j;
                    minimize(dp[i][turn], dp[j - 1][turn - 1] + a[i] - a[j] + 1);
                }
            }
        }

        cout << dp[n][k];
    }
}

namespace subtask2 {
    bool check() {
        return (n <= 5000);
    }

    const int inf = (int) 1e9 + 7;
    const int N = (int) 5e3 + 7;

    int dp[N + 2][N + 2]; // dp[i][j]: tong time khi xet i nguoi dau tien va da bat k lan;
    int f[N + 2];

    void solve() {
        FOR(i, 0, n)
            FOR(j, 0, k) dp[i][j] = inf;

        dp[0][0] = 0;
        f[0] = 0;
        FOR(i, 1, n) f[i] = -a[1];

        FOR(turn, 1, k) {
            FOR(i, 1, n) {
                minimize(dp[i][turn], f[i] + a[i] + 1);
            }

            f[0] = inf;
            FOR(i, 1, n) f[i] = min(f[i - 1], dp[i - 1][turn] - a[i]);
        }

        cout << dp[n][k];
    }
}

namespace subtask3 {
    void solve() {
        vector <int> v;
        FOR(i, 1, n - 1) v.push_back(a[i + 1] - 1 - a[i]);

        sort(v.begin(), v.end(), greater <int> ());
        int ans = a[n] - a[1] + 1;
        FOR(i, 1, k - 1) ans -= v[i - 1];

        cout << ans;
    }
}

int main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin >> n >> k;
    FOR(i, 1, n) cin >> a[i];

    if (subtask1 :: check()) return subtask1 :: solve(), 0;
    if (subtask2 :: check()) return subtask2 :: solve(), 0;
    subtask3 :: solve();

    return 0;
}

• Subtask 1: 20 / 20
• Subtask 2: 30 / 30
• Subtask 3: 50 / 50