#include #include #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 inline bool chmax(T &a, U b) { return a < b and ((a = std::move(b)), true); } template inline bool chmin(T &a, U b) { return a > b and ((a = std::move(b)), true); } template inline int ssize(const T &a) { return (int) std::size(a); } template std::istream &operator>>(std::istream &is, std::vector &a) { for (auto &x: a) is >> x; return is; } template std::ostream &operator<<(std::ostream &os, const std::pair &a) { return os << "(" << a.first << ", " << a.second << ")"; } template 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 struct is_iterable : std::false_type {}; template struct is_iterable())), decltype(std::end(std::declval()))>> : std::true_type { }; template::value && !std::is_same::value && !std::is_same::value>> std::ostream &operator<<(std::ostream &os, const T &a) { return print_seq(a, ", ", "", (os << "{")) << "}"; } void print() { std::cout << "\n"; } template void print(const T &x) { std::cout << x << "\n"; } template void print(const Head &head, Tail... tail) { std::cout << head << " "; print(tail...); } struct Input { template operator T() const { T x; std::cin >> x; return x; } } in; #ifdef MY_DEBUG #include "debug_dump.hpp" #else #define DUMP(...) #endif template 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::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> &adj, const int start) { const int n = int(adj.size()); std::vector 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> bg(n); atcoder::mcf_graph g(n); vector a(n); i64 offset = 0; vector> 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(-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); }