제출 #420202

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
420202		rama_pang	Financial Report (JOI21_financial)	C++17	100 / 100	2166 ms	49520 KiB

Main

#include <bits/stdc++.h>
using namespace std;

struct Segment { // max 0-segment
  bool allZero = false;
  int lft = 0;
  int rgt = 0;
  int opt = 0;

  Segment(int c = 0) {
    if (c == 0) {
      allZero = 1;
      lft = 1;
      rgt = 1;
      opt = 1;
    } else {
      allZero = 0;
      lft = 0;
      rgt = 0;
      opt = 0;
    }
  }

  friend Segment operator+(const Segment &a, const Segment &b) {
    Segment c(0);
    c.allZero = a.allZero && b.allZero;
    c.lft = a.lft + (a.allZero ? b.lft : 0);
    c.rgt = b.rgt + (b.allZero ? a.rgt : 0);
    c.opt = max({a.opt, b.opt, a.rgt + b.lft});
    return c;
  }
};

class MaxSubsegment {
 public:
  int sz;
  vector<Segment> tree;

  MaxSubsegment(int sz) : sz(sz) {
    tree.resize(2 * sz);
    for (int i = sz - 1; i >= 0; i--) {
      tree[i] = tree[i * 2] + tree[i * 2 + 1];
    }
  }

  void Modify(int p, int x) {
    tree[p += sz] = Segment(x);
    for (p /= 2; p > 0; p /= 2) {
      tree[p] = tree[p * 2] + tree[p * 2 + 1];
    }
  }

  Segment Query(int l, int r) {
    Segment lft(1), rgt(1);
    for (l += sz, r += sz + 1; l < r; l /= 2, r /= 2) {
      if (l & 1) lft = lft + tree[l++];
      if (r & 1) rgt = tree[--r] + rgt;
    }
    return lft + rgt;
  }
};

class SegTree {
 public:
  int sz;
  vector<int> tree;

  SegTree(int sz) : sz(sz), tree(2 * sz, -1e9) {}

  int Query(int l, int r) {
    int z = 0;
    for (l += sz, r += sz + 1; l < r; l /= 2, r /= 2) {
      if (l & 1) z = max(z, tree[l++]);
      if (r & 1) z = max(z, tree[--r]);
    }
    return z;
  }

  void Modify(int p, int x) {
    tree[p += sz] = x;
    for (p /= 2; p > 0; p /= 2) {
      tree[p] = max(tree[p * 2], tree[p * 2 + 1]);
    }
  }
};

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);

  int N, D;
  cin >> N >> D;
  vector<int> A(N);
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  { // Coordinate Compression
    vector<int> coords = A;
    sort(begin(coords), end(coords));
    coords.resize(unique(begin(coords), end(coords)) - begin(coords));
    for (auto &a : A) {
      a = lower_bound(begin(coords), end(coords), a) - begin(coords);
    }
  }

  // dp[pos][max_element]
  // find maximum value on dp[pos-D...pos][0...A[pos] - 1] -> dp[pos][A[pos]] = 1 + best
  // for x >= A[pos]: dp[pos-D...pos][x] -> dp[pos][x]
  //
  // If we group all occurrences of value val which has distance <= D, then
  // dp[i][val] <= dp[j][val] for i < j (since we can retake "val").
  // We can transform the sequence such that this condition is fullfilled without
  // the distance <= D condition.
  //
  // How? We scan from least valued element, then when we arrive at "val", group them
  // which has distance <= D. Afterwards, we reassign values s.t. the ones in the back
  // has lower value than the ones in the front.
  //
  // After this transformation, dp[i][val] <= dp[j][val] for i < j. So, we only need
  // to store last occurrence of dp[i][val]. This yields O(N^2).
  //
  // For a value "val", since all adjacent occurrence has distance <= D, then the value
  // "val" is valid only for some subsegment of the array. We can precompute this, then
  // we only need to consider "active" values when transitioning. Afterwards, we don't
  // need the array last[], and it's basically a simple DP optimized with segment tree.
  //
  // Time complexity: O(N log N).

  { // Transformation s.t. dp[i][val] <= dp[j][val] for i < j
    int value = 0;
    vector<vector<int>> occ(N);
    for (int i = 0; i < N; i++) {
      occ[A[i]].emplace_back(i);
    }
    for (int v = 0; v < N; v++) {
      int cnt = 0;
      for (int i = 0; i < int(occ[v].size()); i++) {
        int j = i;
        while (j + 1 < int(occ[v].size()) && occ[v][j + 1] - occ[v][j] <= D) {
          j += 1;
        }
        cnt += 1;
        i = j;
      }
      value += cnt;
      cnt = 0;
      for (int i = 0; i < int(occ[v].size()); i++) {
        int j = i;
        while (j + 1 < int(occ[v].size()) && occ[v][j + 1] - occ[v][j] <= D) {
          j += 1;
        }
        cnt += 1;
        for (int k = i; k <= j; k++) {
          A[occ[v][k]] = value - cnt;
        }
        i = j;
      }
    }
    assert(*max_element(begin(A), end(A)) < N);
  }

  const auto DP_N2_Old = [&](int N, int D, vector<int> A) {
    vector<int> dp(N, -1e9);
    vector<int> last(N, -1e9);
    for (int pos = 0; pos < N; pos++) {
      for (int i = A[pos]; i < N; i++) {
        if (pos - last[i] <= D) {
          last[i] = pos;
        }
      }
      last[A[pos]] = pos;
      dp[A[pos]] = max(dp[A[pos]], 1);
      for (int i = 0; i < A[pos]; i++) {
        if (pos - last[i] <= D) {
          dp[A[pos]] = max(dp[A[pos]], dp[i] + 1);
        }
      }
    }

    int ans = 0;
    for (int i = A[N - 1]; i < N; i++) {
      if (N - 1 - last[i] <= D) {
        ans = max(ans, dp[i]);
      }
    }
    return ans;
  };

  const auto Solve = [&](int N, int D, vector<int> A) {
    vector<int> startValid(N, -1);
    vector<int> endValid(N, -1);
    for (int i = 0; i < N; i++) {
      if (startValid[A[i]] == -1) {
        startValid[A[i]] = i;
      }
      endValid[A[i]] = i;
    }

    vector<vector<int>> occ(N);
    for (int i = 0; i < N; i++) {
      occ[A[i]].emplace_back(i);
    }

    MaxSubsegment maxSeg(N);
    for (int i = 0; i < N; i++) if (endValid[i] != -1) {
      for (auto j : occ[i]) {
        maxSeg.Modify(j, 1);
      }

      int lo = endValid[i];
      int hi = N - 1;
      while (lo < hi) {
        int md = (lo + hi) / 2;
        if (maxSeg.Query(endValid[i], md).opt >= D) {
          hi = md;
        } else {
          lo = md + 1;
        }
      }
      endValid[i] = lo;
    }

    SegTree dpSeg(N);
    vector<vector<int>> endEvent(N + 1);
    for (int i = 0; i < N; i++) {
      endEvent[endValid[i] + 1].emplace_back(i);
    }
    for (int pos = 0; pos < N; pos++) {
      for (auto j : endEvent[pos]) {
        dpSeg.Modify(j, -1e9);
      }
      int val = max({dpSeg.Query(A[pos], A[pos]), 1, 1 + dpSeg.Query(0, A[pos] - 1)});
      dpSeg.Modify(A[pos], val);
    }

    return dpSeg.Query(0, N - 1);
  };

  cout << Solve(N, D, A) << '\n';
  return 0;
}

컴파일 시 표준 에러 (stderr) 메시지

Main.cpp: In function 'int main()':
Main.cpp:162:14: warning: variable 'DP_N2_Old' set but not used [-Wunused-but-set-variable]
  162 |   const auto DP_N2_Old = [&](int N, int D, vector<int> A) {
      |              ^~~~~~~~~

• Subtask 1: 14 / 14
• Subtask 2: 14 / 14
• Subtask 3: 20 / 20
• Subtask 4: 12 / 12
• Subtask 5: 5 / 5
• Subtask 6: 35 / 35