結果

問題 No.2495 Three Sets
ユーザー SSRS
提出日時 2023-10-06 22:52:49
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,607 ms / 3,000 ms
コード長 3,959 bytes
コンパイル時間 1,975 ms
コンパイル使用メモリ 210,176 KB
最終ジャッジ日時 2025-02-17 05:26:32
ジャッジサーバーID
（参考情報） 		judge1 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 18
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
const int MAX = 3000;
const long long INF = 1000000000000000000;
template <typename T>
struct li_chao_tree{
  struct line{
    T a, b;
    line(): a(0), b(INF){
    }
    line(T a, T b): a(a), b(b){
    }
    T get(T x){
      return a * x + b;
    }
  };
  int N;
  vector<T> x;
  vector<line> ST;
  li_chao_tree(const vector<T> &x2){
    x = x2;
    sort(x.begin(), x.end());
    int N2 = x.size();
    N = 1;
    while (N < N2){
      N *= 2;
    }
    x.resize(N);
    for (int i = N2; i < N; i++){
      x[i] = x[N2 - 1];
    }
    ST = vector<line>(N * 2 - 1);
  }
  void line_add(line L, int i, int l, int r){
    T la = L.get(x[l]);
    T lb = ST[i].get(x[l]);
    T ra = L.get(x[r - 1]);
    T rb = ST[i].get(x[r - 1]);
    if (la >= lb && ra >= rb){
      return;
    } else if (la <= lb && ra <= rb){
      ST[i] = L;
    } else {
      int m = (l + r) / 2;
      T ma = L.get(x[m]);
      T mb = ST[i].get(x[m]);
      if (ma < mb){
        swap(L, ST[i]);
        swap(la, lb);
        swap(ra, rb);
      }
      if (la < lb){
        line_add(L, i * 2 + 1, l, m);
      }
      if (ra < rb){
        line_add(L, i * 2 + 2, m, r);
      }
    }
  }
  void line_add(T a, T b){
    line_add(line(a, b), 0, 0, N);
  }
  T get(T x2){
    int p = lower_bound(x.begin(), x.end(), x2) - x.begin();
    p += N - 1;
    T ans = INF;
    ans = min(ans, ST[p].get(x2));
    while (p > 0){
      p = (p - 1) / 2;
      ans = min(ans, ST[p].get(x2));
    }
    return ans;
  }
};
int main(){
  vector<int> N(3);
  for (int i = 0; i < 3; i++){
    cin >> N[i];
  }
  vector<vector<int>> A(3);
  for (int i = 0; i < 3; i++){
    A[i].resize(N[i]);
    for (int j = 0; j < N[i]; j++){
      cin >> A[i][j];
    }
  }
  vector<vector<int>> B(3, vector<int>(MAX * 2 + 1, 0));
  for (int i = 0; i < 3; i++){
    for (int j = 0; j < N[i]; j++){
      B[i][A[i][j] + MAX]++;
    }
  }
  int M = MAX * 2 + 2;
  vector<vector<long long>> X(3, vector<long long>(M, 0));
  vector<vector<long long>> Y(3, vector<long long>(M, 0));
  for (int i = 0; i < 3; i++){
    for (int j = 0; j <= MAX * 2; j++){
      X[i][j + 1] = X[i][j] + B[i][MAX * 2 - j];
      Y[i][j + 1] = Y[i][j] + B[i][MAX * 2 - j] * (MAX - j);
    }
  }
  vector<int> p(M);
  for (int i = 0; i < M; i++){
    p[i] = i;
  }
  sort(p.begin(), p.end(), [&](int i, int j){
    return Y[1][i] < Y[1][j];
  });
  auto cross = [](pair<long long, long long> L1, pair<long long, long long> L2){
    if (L1.second - L2.second >= 0){
      return (L1.second - L2.second) / (L2.first - L1.first);
    } else {
      return (L1.second - L2.second - (L2.first - L1.first - 1)) / (L2.first - L1.first);
    }
  };
  long long ans = 0;
  for (int i = 0; i < M; i++){
    vector<pair<long long, long long>> L(M);
    for (int j = 0; j < M; j++){
      L[j] = make_pair(Y[1][p[j]], Y[0][i] * X[1][p[j]]);
    }
    vector<pair<long long, long long>> L2;
    L2.push_back(L[0]);
    for (int j = 1; j < M; j++){
      if (L[j].first == L2.back().first){
        if (L[j].second > L2.back().second){
          L2.back() = L[j];
        }
      } else {
        while (L2.size() >= 2){
          pair<long long, long long> c1 = L2[L2.size() - 2];
          pair<long long, long long> c2 = L2[L2.size() - 1];
          pair<long long, long long> c3 = L[j];
          if (cross(c1, c2) > cross(c2, c3)){
            L2.pop_back();
          } else {
            break;
          }
        }
        L2.push_back(L[j]);
      }
    }
    int cnt = L2.size();
    int idx = 0;
    for (int j = 0; j < M; j++){
      while (idx < cnt - 1){
        long long x = L2[idx].first * X[2][j] + L2[idx].second;
        long long y = L2[idx + 1].first * X[2][j] + L2[idx + 1].second;
        if (x < y){
          idx++;
        } else {
          break;
        }
      }
      ans = max(ans, L2[idx].first * X[2][j] + L2[idx].second + Y[2][j] * X[0][i]);
    }
  }
  cout << ans << endl;
}
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