$\newcommand{\O}{\mathrm{O}}$
(すべての辺の容量および流すフローの流量が非負整数の場合における)最小費用流を求めるアルゴリズム.
(関数)
solve$(s,t,f)$ : $s$ から $t$ へ流量 $f$ 流すのにかかる最小費用
時間計算量: $\O (|F| m \log n)$ (流量を $|F|$ とする)
template<typename CapType, typename CostType> class MinCostFlow { public: using Cat = CapType; using Cot = CostType; using pti = pair<Cot, int>; struct edge { int to; Cat cap; Cot cost; int rev; }; const int V; const Cot inf; vector<vector<edge> > G; vector<Cot> h, dist; vector<int> prevv, preve; MinCostFlow(int node_size) : V(node_size), inf(numeric_limits<Cot>::max()), G(V), h(V, 0), dist(V), prevv(V), preve(V){} void add_edge(int from, int to, Cat cap, Cot cost){ G[from].push_back((edge){to, cap, cost, (int)G[to].size()}); G[to].push_back((edge){from, 0, -cost, (int)G[from].size() - 1}); } Cot solve(int s, int t, Cat f){ Cot res = 0; while(f > 0){ priority_queue<pti,vector<pti>,greater<pti> > que; fill(dist.begin(), dist.end(), inf); dist[s] = 0; que.push(pti(0, s)); while(!que.empty()){ pti p = que.top(); que.pop(); int v = p.second; if(dist[v] < p.first) continue; for(int i = 0; i < (int)G[v].size(); i++){ edge& e = G[v][i]; if(e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]){ dist[e.to] = dist[v] + e.cost + h[v] - h[e.to]; prevv[e.to] = v, preve[e.to] = i; que.push(pti(dist[e.to], e.to)); } } } if(dist[t] == inf) return -1; for(int i = 0; i < V; i++){ if(dist[i] != inf) h[i] += dist[i]; } Cat d = f; for(int v = t; v != s; v = prevv[v]){ d = min(d, G[prevv[v]][preve[v]].cap); } f -= d; res += h[t] * d; for(int v = t; v != s; v = prevv[v]){ edge& e = G[prevv[v]][preve[v]]; e.cap -= d; G[v][e.rev].cap += d; } } return res; } };
AOJ : Minimum Cost Flow 提出コード