MinCostMaxFlow
Min-cost max-flow. If costs can be negative, call setpi before maxflow, but note that negative cost cycles are not supported. To obtain the actual flow, look at positive values only.
Time: O(F E log(V)) where F is max flow. O(VE) for setpi. 79 lines Tested on kattis:mincostmaxflow, stress-tested against another implementation
content/graph/MinCostMaxFlow.h — Stanford, source: Stanford Notebook
#include <bits/extc++.h>
const ll INF = numeric_limits<ll>::max() / 4;
struct MCMF {
struct edge {
int from, to, rev;
ll cap, cost, flow;
};
int N;
vector<vector<edge>> ed;
vi seen;
vector<ll> dist, pi;
vector<edge*> par;
MCMF(int N) : N(N), ed(N), seen(N), dist(N), pi(N), par(N) {}
void addEdge(int from, int to, ll cap, ll cost) {
if (from == to) return;
ed[from].push_back(edge{ from,to,sz(ed[to]),cap,cost,0 });
ed[to].push_back(edge{ to,from,sz(ed[from])-1,0,-cost,0 });
}
void path(int s) {
fill(all(seen), 0);
fill(all(dist), INF);
dist[s] = 0; ll di;
__gnu_pbds::priority_queue<pair<ll, int>> q;
vector<decltype(q)::point_iterator> its(N);
q.push({ 0, s });
while (!q.empty()) {
s = q.top().second; q.pop();
seen[s] = 1; di = dist[s] + pi[s];
for (edge& e : ed[s]) if (!seen[e.to]) {
ll val = di - pi[e.to] + e.cost;
if (e.cap - e.flow > 0 && val < dist[e.to]) {
dist[e.to] = val;
par[e.to] = &e;
if (its[e.to] == q.end())
its[e.to] = q.push({ -dist[e.to], e.to });
else
q.modify(its[e.to], { -dist[e.to], e.to });
}
}
}
rep(i,0,N) pi[i] = min(pi[i] + dist[i], INF);
}
pair<ll, ll> maxflow(int s, int t) {
ll totflow = 0, totcost = 0;
while (path(s), seen[t]) {
ll fl = INF;
for (edge* x = par[t]; x; x = par[x->from])
fl = min(fl, x->cap - x->flow);
totflow += fl;
for (edge* x = par[t]; x; x = par[x->from]) {
x->flow += fl;
ed[x->to][x->rev].flow -= fl;
}
}
rep(i,0,N) for(edge& e : ed[i]) totcost += e.cost * e.flow;
return {totflow, totcost/2};
}
// If some costs can be negative, call this before maxflow:
void setpi(int s) { // (otherwise, leave this out)
fill(all(pi), INF); pi[s] = 0;
int it = N, ch = 1; ll v;
while (ch-- && it--)
rep(i,0,N) if (pi[i] != INF)
for (edge& e : ed[i]) if (e.cap)
if ((v = pi[i] + e.cost) < pi[e.to])
pi[e.to] = v, ch = 1;
assert(it >= 0); // negative cost cycle
}
};