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  1. #include <bits/stdc++.h>
  2. using namespace std;
  3.  
  4. int main() {
  5.     ios::sync_with_stdio(false);
  6.     cin.tie(nullptr);
  7.  
  8.     int n, m;
  9.     if (!(cin >> n >> m)) return 0;
  10.     vector<int> l(n);
  11.     for (int &x : l) cin >> x;
  12.     vector<int> s(n);
  13.     for (int &x : s) cin >> x;
  14.  
  15.     vector<long long> A(m + 1);
  16.     for (int i = 1; i <= m; ++i) cin >> A[i];
  17.  
  18.     int L = n + m;                     // maximal possible grade
  19.     vector<long long> B(L + 2, 0);     // B[1] is unused, B[L+1] = 0
  20.     for (int i = 1; i <= L; ++i) cin >> B[i];
  21.     vector<long long> D(L + 1);
  22.     for (int i = 1; i <= L; ++i) cin >> D[i];
  23.  
  24.     /* profit of each crystal = A[grade] - cost */
  25.     vector<vector<long long>> profit(L + 1);
  26.     for (int i = 0; i < n; ++i) {
  27.         int g = l[i];                         // 1 … m
  28.         long long p = A[g] - (long long)s[i];
  29.         profit[g].push_back(p);
  30.     }
  31.  
  32.     /* prefix sums of the best k profits for every grade */
  33.     vector<vector<long long>> pref(L + 1);
  34.     for (int v = 1; v <= L; ++v) {
  35.         auto &vec = profit[v];
  36.         sort(vec.begin(), vec.end(), greater<long long>());
  37.         pref[v].assign(vec.size() + 1, 0);
  38.         for (size_t k = 1; k <= vec.size(); ++k)
  39.             pref[v][k] = pref[v][k - 1] + vec[k - 1];
  40.     }
  41.  
  42.     const long long NEG = -(1LL << 60);
  43.     vector<long long> dp(n + 1, NEG), nxt(n + 1, NEG);
  44.     dp[0] = 0;                               // nothing chosen yet
  45.  
  46.     for (int v = 1; v <= L; ++v) {
  47.         int cnt = (int)pref[v].size() - 1;   // number of crystals of this grade
  48.         long long Bnext = (v < L) ? B[v + 1] : 0LL;
  49.         fill(nxt.begin(), nxt.end(), NEG);
  50.         for (int c = 0; c <= n; ++c) if (dp[c] != NEG) {
  51.             for (int k = 0; k <= cnt; ++k) {
  52.                 int total = c + k;
  53.                 int nc = total >> 1;                     // new carry
  54.                 long long gain = pref[v][k];
  55.                 if (total & 1) gain += D[v];
  56.                 gain += (long long)nc * Bnext;
  57.                 long long cand = dp[c] + gain;
  58.                 if (cand > nxt[nc]) nxt[nc] = cand;
  59.             }
  60.         }
  61.         dp.swap(nxt);
  62.     }
  63.  
  64.     long long answer = dp[0];
  65.     if (answer < 0) answer = 0;            // empty set is always possible
  66.     cout << answer << '\n';
  67.     return 0;
  68. }
Success #stdin #stdout 0.01s 5284KB
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https://ideone.com/5CDAD1
language:
C++14 (gcc 8.3)
created:
9 hours ago
visibility:
public

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