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pb_jiang

CF1409E AC

Jan 16th, 2023
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C++ 1.79 KB | None | 0 0
  1. #include <assert.h>
  2. #include <bits/stdc++.h>
  3. using namespace std;
  4. #define dbg(...) logger(#__VA_ARGS__, __VA_ARGS__)
  5. template <typename... Args> void logger(string vars, Args &&... values)
  6. {
  7.     cerr << vars << " = ";
  8.     string delim = "";
  9.     (..., (cerr << delim << values, delim = ", "));
  10.     cerr << endl;
  11. }
  12.  
  13. template <class T> inline auto vv(int m) { return vector<vector<T>>(m, vector<T>(m)); }
  14. template <class T> inline auto vv(int m, int n) { return vector<vector<T>>(m, vector<T>(n)); }
  15. template <class T, T init> inline auto vv(int m) { return vector<vector<T>>(m, vector<T>(m, init)); }
  16. template <class T, T init> inline auto vv(int m, int n) { return vector<vector<T>>(m, vector<T>(n, init)); }
  17.  
  18. template <class T> using mpq = priority_queue<T, vector<T>, greater<T>>;
  19.  
  20. using ll = long long;
  21. using pii = pair<int, int>;
  22. using vl = vector<ll>;
  23. using vi = vector<int>;
  24.  
  25. int main(int argc, char **argv)
  26. {
  27.     int t;
  28.     cin >> t;
  29.     while (t--) {
  30.         int n, k;
  31.         cin >> n >> k;
  32.         vi a(n), b(n);
  33.         for (auto &x : a)
  34.             cin >> x;
  35.         for (auto &x : b)
  36.             cin >> x;
  37.  
  38.         sort(a.begin(), a.end());
  39.         vi one_pass(n);
  40.         vi max_op(n);
  41.         for (int i = 0; i < n; ++i)
  42.             one_pass[i] = upper_bound(a.begin(), a.end(), a[i]) - lower_bound(a.begin(), a.end(), a[i] - k);
  43.         max_op[0] = one_pass[0];
  44.         for (int i = 1; i < n; ++i)
  45.             max_op[i] = max(max_op[i - 1], one_pass[i]);
  46.         int ans = 0;
  47.         for (int i = 0; i < n; ++i) {
  48.             int op_id = lower_bound(a.begin(), a.end(), a[i] - k) - a.begin() - 1;
  49.             int best = (op_id >= 0 ? max_op[op_id] : 0) + one_pass[i];
  50.             ans = max(ans, best);
  51.         }
  52.         cout << ans << endl;
  53.     }
  54.     return 0;
  55. };
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