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CF1379C AC

Mar 10th, 2023
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  1. // Problem: C. Choosing flowers
  2. // Contest: Codeforces - Codeforces Round 657 (Div. 2)
  3. // URL: https://codeforces.com/problemset/problem/1379/C
  4. // Memory Limit: 512 MB
  5. // Time Limit: 1000 ms
  6. //
  7. // Powered by CP Editor (https://cpeditor.org)
  8.  
  9. #include <assert.h>
  10. #include <bits/stdc++.h>
  11. using namespace std;
  12. #define dbg(...) logger(#__VA_ARGS__, __VA_ARGS__)
  13. template <typename... Args> void logger(string vars, Args &&... values)
  14. {
  15.     cerr << vars << " = ";
  16.     string delim = "";
  17.     (..., (cerr << delim << values, delim = ", "));
  18.     cerr << endl;
  19. }
  20.  
  21. template <class T> inline auto vv(int m) { return vector<vector<T>>(m, vector<T>(m)); }
  22. template <class T> inline auto vv(int m, int n) { return vector<vector<T>>(m, vector<T>(n)); }
  23. template <class T, T init> inline auto vv(int m) { return vector<vector<T>>(m, vector<T>(m, init)); }
  24. template <class T, T init> inline auto vv(int m, int n) { return vector<vector<T>>(m, vector<T>(n, init)); }
  25.  
  26. template <class T> using mpq = priority_queue<T, vector<T>, greater<T>>;
  27.  
  28. using ll = long long;
  29. using pii = pair<int, int>;
  30. using vl = vector<ll>;
  31. using vi = vector<int>;
  32. using a2l = array<ll, 2>;
  33.  
  34. int t, n, m;
  35.  
  36. int main(int argc, char **argv)
  37. {
  38.     cin >> t;
  39.     while (t--) {
  40.         cin >> n >> m;
  41.         vector<a2l> vr(m);
  42.         vector<ll> vl(m);
  43.         dbg(n, m);
  44.         for (int i = 0; i < m; ++i)
  45.             cin >> vr[i][1] >> vr[i][0], vl[i] = vr[i][1];
  46.         sort(vr.rbegin(), vr.rend()), sort(vl.begin(), vl.end());
  47.         ll acc = 0, ans = 0;
  48.         for (int i = 0, j = m - 1; i < m; ++i) {
  49.             const auto &rcd = vr[i];
  50.             // dbg(j, n, m - 1 - j, rcd[0], vl[j]);
  51.             while (j >= 0 && n > m - 1 - j && rcd[0] < vl[j])
  52.                 acc += vl[j], --j;
  53.             // dbg(rcd[0], i, vl[j], j, acc, ans);
  54.             if (n == m - 1 - j)
  55.                 ans = max(ans, acc);
  56.             if (n - m + j >= 0)
  57.                 if (rcd[1] > rcd[0])
  58.                     ans = max(ans, acc + (n - m + j + 1) * rcd[0]);
  59.                 else
  60.                     ans = max(ans, acc + rcd[1] + (n - m + j) * rcd[0]);
  61.         }
  62.         cout << ans << endl;
  63.     }
  64.     return 0;
  65. };
  66.  
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