제출 #1086474

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1086474		vjudge1	Fireworks (APIO16_fireworks)	C++17	100 / 100	207 ms	102740 KiB

fireworks

#include <iostream>
#include <map>
#include <tuple>
#include <utility>
#include <vector>

struct ConvexLinearSet {
  int m = 0;
  long long c = 0;
  std::map<long long, int> m_changes; // key: xpos, val: dm

  int size() {
    return m_changes.size();
  }
  
  void merge_with(const ConvexLinearSet& o) {
    m += o.m;
    c += o.c;
    for (auto [x, dm]: o.m_changes) {
      m_changes[x] += dm;
    }
  }

  std::tuple<long long, long long, int> pop_until_negative_gradient() {
    long long x, x_prev;
    int dm;
    while (!m_changes.empty() && m >= 0) {
      // pop largest
      x_prev = x;
      std::tie(x, dm) = *m_changes.rbegin();
      m_changes.erase(x);

      // update current m & c
      // m*x + c = (m - dm)*x + new_c -> new_c = x*dm + c
      m -= dm;
      c += x * dm;
    }

    return {x_prev, x, dm};
  }

  long long query_min() {
    long long x, x_prev;
    int dm;
    std::tie(x_prev, x, dm) = pop_until_negative_gradient();
    return x * m + c;
  }

  void apply_w(int w) {
    long long x, x_prev;
    int dm;
    std::tie(x_prev, x, dm) = pop_until_negative_gradient();

    // x = leftmost xpos that make the minimum
    long long minval = x * m + c;

    if (m != -1) {
      m_changes[x] -= m + 1; // make gradient = -1 at x
    }
    m_changes[x + w]++; // make next gradient = 0 at x + w

    // we are at m = 0 now, jump to the end of m = 0
    x = m == -dm ? x_prev : x;
    x += w;

    // make gradient = 1
    // const = x + new_c -> new_c = const - x
    m_changes[x]++;
    m = 1;
    c = minval - x;
  }
};

int main() {
  std::cin.tie(NULL)->sync_with_stdio(false);

  int n, m;
  std::cin >> n >> m;

  std::vector<ConvexLinearSet> sets(n + m + 1);
  std::vector<std::vector<int>> adj(n + m + 1);
  std::vector<int> w(n + m + 1);
  w[1] = 0;
  for (int i = 2; i <= n + m; i++) {
    int par;
    std::cin >> par >> w[i];

    adj[par].push_back(i);
    if (i > n) {
      // leaf nodes
      sets[i].m = 1;
      sets[i].c = -w[i];
      sets[i].m_changes[w[i]] = 2;
    }
  }

  // idea: dp[u][x] = apply_w(sum(dp[v][x]), w[u])
  //      dp[leaf][x] = abs(x - w[leaf])
  // apply_w with convex linear func -> we know these two greedy logic:
  //   1. for gradient > 1, we can instead increase from w => we can have gradient=1 instead
  //   2. for gradient < 0, we can use up all w first => x..x+w will have gradient=-1
  // so at minimum point (mx, my), we can have line (mx, my + w) to (mx + c, my) and then (mx + c, my) to (last_mx + c, my), then gradient=1 afterwards
  // but why shift right? because we can think of all the base constant are initially moved by w, and later on we cut at minimum points and apply (2) to the left direction.

  auto dfs = [&](auto &&dfs_fn, int u) -> void {
    if (adj[u].empty()) {
      return;
    }

    // smaller to larger trick here
    // each leaf element will only be merged at most logN times
    for (int v: adj[u]) {
      dfs_fn(dfs_fn, v);
      if (sets[u].size() < sets[v].size()) {
        // find largest set
        std::swap(sets[u], sets[v]);
      }
    }

    // merge sets to the largest one
    for (int v: adj[u]) {
      sets[u].merge_with(sets[v]);
    }

    sets[u].apply_w(w[u]);
  };

  dfs(dfs, 1);
  std::cout << sets[1].query_min() << std::endl;

  return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 19 / 19
• Subtask 3: 29 / 29
• Subtask 4: 45 / 45