結果

問題 No.1678 Coin Trade (Multiple)
コンテスト
ユーザー keijak
提出日時 2021-09-13 13:22:08
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 5,148 bytes
コンパイル時間 3,269 ms
コンパイル使用メモリ 227,204 KB
最終ジャッジ日時 2025-01-24 13:35:24
ジャッジサーバーID
（参考情報） 		judge4 / judge2
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 * 35 TLE * 21
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 # 

#include <bits/stdc++.h>
#include <atcoder/mincostflow>
#define REP_(i, a_, b_, a, b, ...) \
  for (int i = (a), END_##i = (b); i < END_##i; ++i)
#define REP(i, ...) REP_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
#define ALL(x) std::begin(x), std::end(x)
using i64 = long long;

template<typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b and ((a = std::move(b)), true);
}
template<typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b and ((a = std::move(b)), true);
}
template<typename T>
inline int ssize(const T &a) {
  return (int) std::size(a);
}

template<typename T>
std::istream &operator>>(std::istream &is, std::vector<T> &a) {
  for (auto &x: a) is >> x;
  return is;
}
template<typename T, typename U>
std::ostream &operator<<(std::ostream &os, const std::pair<T, U> &a) {
  return os << "(" << a.first << ", " << a.second << ")";
}
template<typename Container>
std::ostream &print_seq(const Container &a, std::string_view sep = " ",
                        std::string_view ends = "\n",
                        std::ostream &os = std::cout) {
  auto b = std::begin(a), e = std::end(a);
  for (auto it = std::begin(a); it != e; ++it) {
    if (it != b) os << sep;
    os << *it;
  }
  return os << ends;
}
template<typename T, typename = void>
struct is_iterable : std::false_type {};
template<typename T>
struct is_iterable<T, std::void_t<decltype(std::begin(std::declval<T>())),
                                  decltype(std::end(std::declval<T>()))>>
    : std::true_type {
};

template<typename T, typename = std::enable_if_t<
    is_iterable<T>::value &&
        !std::is_same<T, std::string_view>::value &&
        !std::is_same<T, std::string>::value>>
std::ostream &operator<<(std::ostream &os, const T &a) {
  return print_seq(a, ", ", "", (os << "{")) << "}";
}

void print() { std::cout << "\n"; }
template<class T>
void print(const T &x) {
  std::cout << x << "\n";
}
template<typename Head, typename... Tail>
void print(const Head &head, Tail... tail) {
  std::cout << head << " ";
  print(tail...);
}

struct Input {
  template<typename T>
  operator T() const {
    T x;
    std::cin >> x;
    return x;
  }
} in;

#ifdef MY_DEBUG
#include "debug_dump.hpp"
#else
#define DUMP(...)
#endif

template<class T>
T ceil_div(T x, T y) {
  assert(y != 0);
  return x / y + (((x ^ y) >= 0) and (x % y));
}

const i64 INF = std::numeric_limits<i64>::max();

struct Edge {
  i64 cost;
  int to;
};

// Finds the shortest path from the start node and detects negative cycle if
// exists. Returns the minimum cost from the start node to each node. INF
// indicates unreachable. -INF indicates having a negative cycle in a path from
// the start node.
auto bellman_ford(const std::vector<std::vector<Edge>> &adj, const int start) {
  const int n = int(adj.size());
  std::vector<i64> mincost(n, INF);
  mincost[start] = 0;

  for (int k = 0; k < n - 1; ++k) {
    for (int i = 0; i < n; ++i) {
      const i64 di = mincost[i];
      if (di == INF) continue;  // Haven't reached i yet.
      for (const Edge &e: adj[i]) {
        if (mincost[e.to] > di + e.cost) {
          mincost[e.to] = di + e.cost;
        }
      }
    }
  }

  // Negative cycle detection.
  // If there's no negative cycle, at least one node gets the shortest
  // distance determined in each iteration above. If we have gone through N-1
  // iterations and still have an update, there must be a negative cycle.
  for (int k = 0; k < n; ++k) {
    bool updated = false;
    for (int i = 0; i < n; ++i) {
      const i64 di = mincost[i];
      if (di == INF) continue;
      const bool in_negative_cycle = (di == -INF);
      for (const Edge &e: adj[i]) {
        if (mincost[e.to] == -INF) continue;
        if (in_negative_cycle or (mincost[e.to] > di + e.cost)) {
          mincost[e.to] = -INF;
          updated = true;
        }
      }
    }
    if (not updated) break;
  }

  return mincost;
}

using namespace std;

auto solve() {
  const int n = in, K = in;
  vector<vector<Edge>> bg(n);
  atcoder::mcf_graph<int, i64> g(n);
  vector<i64> a(n);
  i64 offset = 0;
  vector<tuple<i64, int, int>> edges;
  REP(i, n) {
    a[i] = in;
    int m = in;
    REP(j, m) {
      int b = in;
      --b;
      i64 cost = a[b] - a[i];
      if (cost < 0) {
        bg[b].push_back({cost, i});
        chmax(offset, ceil_div<i64>(-cost, i - b));
        edges.emplace_back(cost, b, i);
      }
    }
  }
  REP(i, n - 1) {
    bg[i].push_back({0, i + 1});
  }
  auto potential = bellman_ford(bg, 0);
  assert (all_of(ALL(potential), [](i64 x) { return x < INF; }));
  DUMP(potential);

  for (auto[cost, from, to]: edges) {
    assert(from < to);
    assert(cost < 0);
    assert(cost + offset * (to - from) >= 0);
    g.add_edge(from, to, 1, cost + (potential[from] - potential[to]));
  }
  REP(i, n - 1) {
    g.add_edge(i, i + 1, K, potential[i] - potential[i + 1]);
  }
  auto[mf, mc] = g.flow(0, n - 1, K);
  assert(mf == K);
  DUMP(mc);
  DUMP(offset);
  mc -= K * (potential[0] - potential[n - 1]);
  return -mc;
}

int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  auto ans = solve();
  print(ans);
}
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