Submission #1272805

#	Time	Username	Problem	Language	Result	Execution time	Memory
1272805		witek_cpp	Let's Win the Election (JOI22_ho_t3)	C++20	21 / 100	1230 ms	8384 KiB

Main

#include <iostream>
#include <vector>
#include <algorithm>
#include <iomanip>

using namespace std;
typedef long long ll;
typedef long double ld;

#define st first
#define nd second
#define f(a, c, b) for (int a = c; b > a; a++)
#define pb push_back
#define all(a) a.begin(), a.end()
#define wczytaj(a, c, n) a.resize(n); f(i, c, n) cin >> a[i];
#define sz(a) int(a.size())
#define wypisz(a, c) f(i, c, sz(a)) cout << a[i] << " "; cout << "\n";

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    const ld INF = 1'000'000'000'000;
    vector<pair<ll, ll>> a(n);
    f(i, 0, n) cin >> a[i].nd >> a[i].st;
    f(i, 0, n) if (a[i].st == -1) a[i].st = INF;
    sort(all(a));
    vector<ld> A(n + 1);
    vector<ld> B(n + 1);
    f(i, 0, n) A[i + 1] = a[i].nd;
    f(i, 0, n) B[i + 1] = a[i].st;
    vector<vector<ld>> dp_suf(n + 2, vector<ld>(n + 2));
    vector<vector<ld>> dp(n + 1, vector<ld>(n + 1));
    ld ans = INF;
    f(ile, 0, k) {
        f(j, 1, n + 2) f(i, 1, n + 2) dp_suf[i][j] = INF;
        f(i, 1,n + 2) dp_suf[i][0] = 0;
        for (int i = n; i > 0; i--) {
            for (int j = (n - i + 1); j > 0; j--) {
                dp_suf[i][j] = min(dp_suf[i + 1][j], (dp_suf[i + 1][j - 1] + (A[i]/ld(ile + 1))));
            }
        }
        if (ile == 0) {
            ans = min(ans, dp_suf[1][k]);
            continue;
        }
        f(i, 0, n + 1) f(j, 0,n + 1) dp[i][j] = INF;
        dp[0][0] = 0;
        f(i, 1, n + 1) {
            f(j, 1, min(ile, i) + 1) {
                dp[i][j] = min(dp[i][j], dp[i - 1][j] + A[i]/ld(ile + 1));
                dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + (B[i]/ld(j)));
            }
        }
        f(i, 1, n + 1) {
            if (i <= k) ans = min(ans, dp[i][ile] + dp_suf[i + 1][k - i]);
        }
    }
    cout << fixed << setprecision(6) << ans;
    return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 5 / 5
• Subtask 3: 0 / 11
• Subtask 4: 0 / 12
• Subtask 5: 0 / 33
• Subtask 6: 11 / 11
• Subtask 7: 0 / 23