费用流
给定一个带费用的网络，规定 \((u,v)\) 间的费用为 \(f(u,v) \times w(u,v)\) ，求解该网络中总花费最小的最大流称之为最小费用最大流。总时间复杂度为 \(\mathcal O(NMf)\) ，其中 \(f\) 代表最大流。
struct MinCostFlow {using LL = long long;using PII = pair<LL, int>;const LL INF = numeric_limits<LL>::max();struct Edge {int v, c, f;Edge(int v, int c, int f) : v(v), c(c), f(f) {}};const int n;vector<Edge> e;vector<vector<int>> g;vector<LL> h, dis;vector<int> pre;MinCostFlow(int n) : n(n), g(n) {}void add(int u, int v, int c, int f) { // c 流量, f 费用// if (f < 0) {// g[u].push_back(e.size());// e.emplace_back(v, 0, f);// g[v].push_back(e.size());// e.emplace_back(u, c, -f);// } else {g[u].push_back(e.size());e.emplace_back(v, c, f);g[v].push_back(e.size());e.emplace_back(u, 0, -f);// }}bool dijkstra(int s, int t) {dis.assign(n, INF);pre.assign(n, -1);priority_queue<PII, vector<PII>, greater<PII>> que;dis[s] = 0;que.emplace(0, s);while (!que.empty()) {auto [d, u] = que.top();que.pop();if (dis[u] < d) continue;for (int i : g[u]) {auto [v, c, f] = e[i];if (c > 0 && dis[v] > d + h[u] - h[v] + f) {dis[v] = d + h[u] - h[v] + f;pre[v] = i;que.emplace(dis[v], v);}}}return dis[t] != INF;}pair<int, LL> flow(int s, int t) {int flow = 0;LL cost = 0;h.assign(n, 0);while (dijkstra(s, t)) {for (int i = 0; i < n; ++i) h[i] += dis[i];int aug = numeric_limits<int>::max();for (int i = t; i != s; i = e[pre[i] ^ 1].v) aug = min(aug, e[pre[i]].c);for (int i = t; i != s; i = e[pre[i] ^ 1].v) {e[pre[i]].c -= aug;e[pre[i] ^ 1].c += aug;}flow += aug;cost += LL(aug) * h[t];}return {flow, cost};}
};