Submission #885780

#	Time	Username	Problem	Language	Result	Execution time	Memory
885780		EJIC_B_KEDAX	Let's Win the Election (JOI22_ho_t3)	C++14	10 / 100	845 ms	994860 KiB

Main

#ifdef LOCAL
    //#define _GLIBCXX_DEBUG
#endif
#include <bits/stdc++.h>
#include <random>

#ifndef LOCAL
    //#pragma GCC optimize("O3")
    //#pragma GCC optimize("Ofast")
    //#pragma GCC optimize("unroll-loops")
    #pragma GCC target("avx,avx2,bmi,bmi2,popcnt,lzcnt")
#endif
using namespace std;
using ll = long long;
using ld = long double;
#define x first
#define y second
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

mt19937_64 mt(418);

void solve();
void init();

int32_t main() {
#ifndef LOCAL
    cin.tie(nullptr)->sync_with_stdio(false);
#endif
    // srand(time(0));
    init();
    cout << fixed << setprecision(20);
    int t = 1;
    // cin >> t;
    while (t--) {
        solve();
    }
}

void init() {}

int lay = 1;

bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
    if (a.y < b.y) {
        return true;
    }
    if (a.y > b.y) {
        return false;
    }
    return a.x < b.x;
}


void solve() {
    int n, k;
    cin >> n >> k;
    vector<pair<int, int>> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i].x >> a[i].y;
        if (a[i].y == -1) {
            a[i].y = INT32_MAX / 2;
        }
    }
    double ans = 2e18;
    vector<vector<vector<double>>> dp(n + 1, vector<vector<double>>(k + 1, vector<double>(n + 1, 2e18)));
    // for (int ll = 0; ll <= n; ll++) {
        sort(all(a), cmp);
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= k; j++) {
                for (int l = 0; l <= n; l++) {
                    dp[i + 1][j][l] = 2e18;
                }
            }
        }
        dp[0][0][0] = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= k; j++) {
                for (int l = 0; l <= n; l++) {
                    dp[i + 1][j][l] = min(dp[i + 1][j][l], dp[i][j][l]);
                    if (j > 0) {
                        dp[i + 1][j][l] = min(dp[i + 1][j][l], dp[i][j - 1][l] + a[i].x / (double)(l + 1));
                        if (l > 0) {
                            dp[i + 1][j][l] = min(dp[i + 1][j][l], dp[i][j - 1][l - 1] + a[i].y / (double)l);
                        }
                    }
                }
            }
        }
        for (int i = 0; i <= n; i++) {
            ans = min(ans, dp[n][k][i]);
        }
        lay++;
    // }
    cout << ans << '\n';
}

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