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Python

# https://en.wikipedia.org/wiki/Minimum-cost_flow_problem in O(V^2 * E)
from heapq import *
class Edge:
def __init__(self, to, cap, cost, rev):
self.to = to
self.cap = cap
self.cost = cost
self.rev = rev
self.f = 0
def add_edge(graph, s, t, cap, cost):
graph[s].append(Edge(t, cap, cost, len(graph[t])))
graph[t].append(Edge(s, 0, -cost, len(graph[s]) - 1))
def bellman_ford(graph, s, dist):
n = len(graph)
dist[:] = [float('inf')] * n
dist[s] = 0
inqueue = [False] * n
q = [0] * n
qt = 0
q[qt] = s
qt += 1
qh = 0
while (qh - qt) % n != 0:
u = q[qh % n]
inqueue[u] = False
for i in range(len(graph[u])):
e = graph[u][i]
if e.cap <= e.f:
continue
v = e.to
ndist = dist[u] + e.cost
if dist[v] > ndist:
dist[v] = ndist
if not inqueue[v]:
inqueue[v] = True
q[qt % n] = v
qt += 1
qh += 1
def min_cost_flow(graph, s, t, maxf):
n = len(graph)
prio = [0] * n
curflow = [0] * n
prevedge = [0] * n
prevnode = [0] * n
pot = [0] * n
bellman_ford(graph, s, pot) # bellmanFord invocation can be skipped if edges costs are non-negative
flow = 0
flow_cost = 0
while flow < maxf:
q = [(0, s)]
prio[:] = [float('inf')] * n
prio[s] = 0
finished = [False] * n
curflow[s] = float('inf')
while not finished[t] and q:
(prio_u, u) = heappop(q)
if prio_u != prio[u]:
continue
finished[u] = True
for i in range(len(graph[u])):
e = graph[u][i]
if e.f >= e.cap:
continue
v = e.to
nprio = prio[u] + e.cost + pot[u] - pot[v]
if prio[v] > nprio:
prio[v] = nprio
heappush(q, (nprio, v))
prevnode[v] = u
prevedge[v] = i
curflow[v] = min(curflow[u], e.cap - e.f)
if prio[t] == float('inf'):
break
for i in range(n):
if finished[i]:
pot[i] += prio[i] - prio[t]
df = min(curflow[t], maxf - flow)
flow += df
v = t
while v != s:
e = graph[prevnode[v]][prevedge[v]]
e.f += df
graph[v][e.rev].f -= df
flow_cost += df * e.cost
v = prevnode[v]
return flow, flow_cost
def test():
graph = [[] for _ in range(3)]
add_edge(graph, 0, 1, 3, 1)
add_edge(graph, 0, 2, 2, 1)
add_edge(graph, 1, 2, 2, 1)
assert (4, 6) == min_cost_flow(graph, 0, 2, float('inf'))
test()