Submission #593665

#	Time	Username	Problem	Language	Result	Execution time	Memory
593665		piOOE	Let's Win the Election (JOI22_ho_t3)	C++17	56 / 100	109 ms	8284 KiB

Main

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using db = double;

const int N = 101;

db dp[N][N][N]; //id, b, a

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, K;
    cin >> n >> K;
    vector<int> a(n), b(n), oa(n), ob(n);
    int cntb = 0;
    for (int i = 0; i < n; ++i) {
        cin >> a[i] >> b[i];
        if (b[i] != -1) {
            cntb += 1;
        }
    }
    iota(oa.begin(), oa.end(), 0), iota(ob.begin(), ob.end(), 0);
    sort(oa.begin(), oa.end(), [&a](int i, int j) {
        return a[i] < a[j];
    });
    sort(ob.begin(), ob.end(), [&](int i, int j) {
        if (b[i] == -1) return false;
        if (b[j] == -1) return true;
        return b[i] < b[j];
    });
    const db inf = 1e12;
    auto solve = [&](int mx) -> db {
        for (int i = 0; i <= n; ++i) {
            for (int j = 0; j <= mx; ++j) {
                for (int k = 0; k <= K - j; ++k) {
                    dp[i][j][k] = inf;
                }
            }
        }
        dp[0][0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            //don't take
            for (int j = 0; j <= mx; ++j) {
                for (int k = 0; k <= K - j; ++k) {
                    dp[i][j][k] = dp[i - 1][j][k];
                }
            }
            //take a
            int idx = ob[i - 1];
            for (int j = 0; j <= min(i, mx); ++j) {
                for (int k = 1; k <= min(i, K) - j; ++k) {
                    dp[i][j][k] = min(dp[i][j][k], dp[i - 1][j][k - 1] + a[idx] / db(mx + 1));
                }
            }
            if (b[idx] != -1) {
                for (int j = 1; j <= min(i, mx); ++j) {
                    for (int k = 0; k <= min(i, K) - j; ++k) {
                        dp[i][j][k] = min(dp[i][j][k], dp[i - 1][j - 1][k] + b[idx] / db(j));
                    }
                }
            }
        }
        return dp[n][mx][K - mx];
    };
    db ans = inf;
    for (int cnt = 0; cnt <= min(K, cntb); ++cnt) {
        ans = min(ans, solve(cnt));
    }
    cout << fixed << setprecision(20) << ans;
    return 0;
}

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