// TLE解 O(N^2K)本の辺を張り,下駄をはかせる #include using namespace std; using ll = long long; #include #include #include #include #include namespace atcoder { template struct mcf_graph { public: mcf_graph() {} mcf_graph(int n) : _n(n), g(n) {} int add_edge(int from, int to, Cap cap, Cost cost) { assert(0 <= from && from < _n); assert(0 <= to && to < _n); assert(0 <= cap); assert(0 <= cost); int m = int(pos.size()); pos.push_back({from, int(g[from].size())}); int from_id = int(g[from].size()); int to_id = int(g[to].size()); if (from == to) to_id++; g[from].push_back(_edge{to, to_id, cap, cost}); g[to].push_back(_edge{from, from_id, 0, -cost}); return m; } struct edge { int from, to; Cap cap, flow; Cost cost; }; edge get_edge(int i) { int m = int(pos.size()); assert(0 <= i && i < m); auto _e = g[pos[i].first][pos[i].second]; auto _re = g[_e.to][_e.rev]; return edge{ pos[i].first, _e.to, _e.cap + _re.cap, _re.cap, _e.cost, }; } std::vector edges() { int m = int(pos.size()); std::vector result(m); for (int i = 0; i < m; i++) { result[i] = get_edge(i); } return result; } std::pair flow(int s, int t) { return flow(s, t, std::numeric_limits::max()); } std::pair flow(int s, int t, Cap flow_limit) { return slope(s, t, flow_limit).back(); } std::vector> slope(int s, int t) { return slope(s, t, std::numeric_limits::max()); } std::vector> slope(int s, int t, Cap flow_limit) { assert(0 <= s && s < _n); assert(0 <= t && t < _n); assert(s != t); // variants (C = maxcost): // -(n-1)C <= dual[s] <= dual[i] <= dual[t] = 0 // reduced cost (= e.cost + dual[e.from] - dual[e.to]) >= 0 for all edge std::vector dual(_n, 0), dist(_n); std::vector pv(_n), pe(_n); std::vector vis(_n); auto dual_ref = [&]() { std::fill(dist.begin(), dist.end(), std::numeric_limits::max()); std::fill(pv.begin(), pv.end(), -1); std::fill(pe.begin(), pe.end(), -1); std::fill(vis.begin(), vis.end(), false); struct Q { Cost key; int to; bool operator<(Q r) const { return key > r.key; } }; std::priority_queue que; dist[s] = 0; que.push(Q{0, s}); while (!que.empty()) { int v = que.top().to; que.pop(); if (vis[v]) continue; vis[v] = true; if (v == t) break; // dist[v] = shortest(s, v) + dual[s] - dual[v] // dist[v] >= 0 (all reduced cost are positive) // dist[v] <= (n-1)C for (int i = 0; i < int(g[v].size()); i++) { auto e = g[v][i]; if (vis[e.to] || !e.cap) continue; // |-dual[e.to] + dual[v]| <= (n-1)C // cost <= C - -(n-1)C + 0 = nC Cost cost = e.cost - dual[e.to] + dual[v]; if (dist[e.to] - dist[v] > cost) { dist[e.to] = dist[v] + cost; pv[e.to] = v; pe[e.to] = i; que.push(Q{dist[e.to], e.to}); } } } if (!vis[t]) { return false; } for (int v = 0; v < _n; v++) { if (!vis[v]) continue; // dual[v] = dual[v] - dist[t] + dist[v] // = dual[v] - (shortest(s, t) + dual[s] - dual[t]) + (shortest(s, v) + dual[s] - dual[v]) // = - shortest(s, t) + dual[t] + shortest(s, v) // = shortest(s, v) - shortest(s, t) >= 0 - (n-1)C dual[v] -= dist[t] - dist[v]; } return true; }; Cap flow = 0; Cost cost = 0, prev_cost_per_flow = -1; std::vector> result; result.push_back({flow, cost}); while (flow < flow_limit) { if (!dual_ref()) break; Cap c = flow_limit - flow; for (int v = t; v != s; v = pv[v]) { c = std::min(c, g[pv[v]][pe[v]].cap); } for (int v = t; v != s; v = pv[v]) { auto& e = g[pv[v]][pe[v]]; e.cap -= c; g[v][e.rev].cap += c; } Cost d = -dual[s]; flow += c; cost += c * d; if (prev_cost_per_flow == d) { result.pop_back(); } result.push_back({flow, cost}); prev_cost_per_flow = d; } return result; } private: int _n; struct _edge { int to, rev; Cap cap; Cost cost; }; std::vector> pos; std::vector> g; }; } // namespace atcoder using namespace atcoder; #define rep2(i, m, n) for (int i = (m); i < (n); ++i) #define rep(i, n) rep2(i, 0, n) template istream &operator>>(istream &is, vector &v) { for (auto &e : v) is >> e; return is; } struct fast_ios { fast_ios(){ cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_; // 辺 ij の流量を x から x+1 にするときの追加コストは, // c_{ijx} = (x+1 - p_{ij}) - (x - p_{ij})^2 = 2 (x - p_{ij}) + 1 // 辺 ij をコピーして辺 ijx とする. // x は 0 <= x < a_i の範囲しか考えなくてよい. // c_{ijx} の最小値は -300 以上なので,BIG = 300 として c_{ijx} <- c_{ijx} + BIG とゲタを履かせれば,コストが非負になる. // 辺 ijx のうち,使われるものの総数は必ず K なので,最後に費用から K * BIG を引けばよい int main() { const ll BIG = 1000; int n, k; cin >> n >> k; int s = 2*n; int t = 2*n + 1; int N = 2*n + 2; mcf_graph g(N); vector a(n), b(n); cin >> a >> b; rep(i, n) g.add_edge(s, i, a[i], 0); rep(i, n) g.add_edge(n+i, t, b[i], 0); ll ans = 0; rep(i, n) rep(j, n) { ll p; cin >> p; ans += p*p; // rep(x, a[i]) g.add_edge(i, n+j, 1, 2*(x-p) + 1 + BIG); // こっちは AC rep(x, k) g.add_edge(i, n+j, 1, 2*(x-p) + 1 + BIG); // こっちは TLE を期待 } auto [cap, cost] = g.flow(s, t, k); ans += cost - k * BIG; cout << ans << '\n'; return 0; }