Submission #643496

#	Time	Username	Problem	Language	Result	Execution time	Memory
643496		lunchbox1	Let's Win the Election (JOI22_ho_t3)	C++17	100 / 100	108 ms	2388 KiB

Main

/*
sort by b from small to large

optimal construction can be split into two halves
- on the left, all the things are taken
- on the right, only a's are taken

fix the number of b's, let this be t
f(i,j) = min time to use prefix i with j b's
f(i,j) = min(f(i-1,j) + a[i] / t, f(i-1,j-1) + b[i] / t)

now for the suffixes, we need to pick k-i a's at the end,
we must pick the ones with the least a[j] value. this part
can be done in O(n log^2 n)

overall the time complexity is O(n^3)
*/
#include <bits/stdc++.h>
using namespace std;

#ifdef LOCAL
  #include "debug.h"
#else
  #define debug(...) 0
#endif

const int N = 505;
const double INF = 1e18;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  static array<double, 2> p[N];
  for (int i = 0; i < n; i++) {
    double a, b;
    cin >> a >> b;
    if (b == -1)
      b = INF;
    p[i] = {b, a};
  }
  sort(p, p + n);
  static double g[N][N];
  for (int i = 0; i < n; i++) {
    vector<double> t;
    for (int j = i; j < n; j++)
      t.push_back(p[j][1]);
    sort(t.begin(), t.end());
    g[i][0] = 0;
    for (int j = 1; j <= (int) t.size(); j++)
      g[i][j] = g[i][j - 1] + t[j - 1];
  }
  double ans = g[0][k];
  for (int t = 0; t <= k; t++) {
    static double f[N];
    for (int i = 0; i <= t; i++)
      f[i] = INF;
    f[0] = 0;
    for (int i = 0; i < n; i++) {
      for (int j = t; j >= 0; j--)
        f[j] = min(f[j] + p[i][1] / (t + 1), j == 0 ? INF : f[j - 1] + p[i][0] / j);
      if (k - i - 1 >= 0)
        ans = min(ans, f[t] + g[i + 1][k - i - 1] / (t + 1));
    }
  }
  cout << fixed << setprecision(12) << ans << '\n';
  return 0;
}

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