Submission #1199986

#	Time	Username	Problem	Language	Result	Execution time	Memory
1199986		myst6	Cloud Computing (CEOI18_clo)	C++20	0 / 100	89 ms	836 KiB

clo

#include <bits/stdc++.h>

using namespace std;
using ll = long long;

const ll inf = 1LL << 60;

void chmax(ll &a, ll b) { a = max(a, b); }

int main() {
  cin.tie(0)->sync_with_stdio(0);
  
  int n;
  cin >> n;
  vector<int> C(n), F(n), V(n);
  for (int i=0; i<n; i++) cin >> C[i] >> F[i] >> V[i];
  
  int m;
  cin >> m;
  vector<int> qC(m), qF(m), qV(m);
  for (int i=0; i<m; i++) cin >> qC[i] >> qF[i] >> qV[i];

  vector<int> order(n);
  iota(order.begin(), order.end(), 0);
  sort(order.begin(), order.end(), [&](int i, int j) -> bool {
    return F[i] < F[j];
  });

  vector<int> qOrder(m);
  iota(qOrder.begin(), qOrder.end(), 0);
  sort(qOrder.begin(), qOrder.end(), [&](int i, int j) -> bool {
    return qF[i] > qF[j];
  });

  // ill try subtask 4 first
  // dp[i][j]: only consider orders with index <= i
  //    and we have a coresum of j currently, what's the max profit
  // but like we can just do this same thing but we sort q in ascending frequency
  // and only add the cores to the rows which they are allowed on??
  vector dp(m+1, vector<ll>(51, -inf));
  dp[0][0] = 0;
  for (int i=0; i<n; i++) {
    for (int j=0; j<m; j++) {
      int jj = qOrder[j];
      // add core on this row
      if (F[i] >= qF[jj]) {
        for (int k=50-C[i]; k>=0; k--) {
          chmax(dp[j][k+C[i]], dp[j][k] - V[i]);
        }
      }
      // take any subset onto the next row
      for (int k=0; k<=50; k++) {
        chmax(dp[j+1][k], dp[j][k]);
      }
      // take a subset and use it
      for (int k=qC[jj]; k<=50; k++) {
        chmax(dp[j+1][k-qC[jj]], dp[j][k] + qV[jj]);
      }
    }
  }

  // answer can be from any dp cell
  ll ans = -inf;
  for (int j=0; j<=m; j++) {
    for (int k=0; k<=50; k++) {
      chmax(ans, dp[j][k]);
    }
  }

  cout << ans << "\n";

  return 0;
}

• Subtask 1: 0 / 18
• Subtask 2: 0 / 18
• Subtask 3: 0 / 18
• Subtask 4: 0 / 18
• Subtask 5: 0 / 18
• Subtask 6: 0 / 10