概要

各辺\(e\)が容量\(c_e\)と単位コスト\(d_e\)を持つ、有向グラフ\(G = (V, E)\)に対し、 特定のsourceとsinkの間に流量\(F\)を流す時の最小コストを求める。

蟻本[1]を参考にして実装。

with Bellman-Ford Algorithm

以下を、流量\(F\)を流しきるまで繰り返す

  1. Bellman-Ford Algorithmでsourceとsinkの間の最小コストとなるパスを求める

  2. そのパスに流せる限り流す

計算量

\(O(F |V| |E|)\)

実装

# 最小費用流(minimum cost flow)
class MinCostFlow:
    def __init__(self, n):
        self.n = n
        self.G = [[] for i in range(n)]

    def addEdge(self, f, t, cap, cost):
        # [to, cap, cost, rev]
        self.G[f].append([t, cap, cost, len(self.G[t])])
        self.G[t].append([f, 0, -cost, len(self.G[f])-1])

    def minCostFlow(self, s, t, f):
        n = self.n
        G = self.G
        prevv = [0]*n; preve = [0]*n
        INF = 10**9+7

        res = 0
        while f:
            dist = [INF]*n
            dist[s] = 0
            update = 1
            while update:
                update = 0
                for v in range(n):
                    if dist[v] == INF:
                        continue
                    gv = G[v]
                    for i in range(len(gv)):
                        to, cap, cost, rev = gv[i]
                        if cap > 0 and dist[v] + cost < dist[to]:
                            dist[to] = dist[v] + cost
                            prevv[to] = v; preve[to] = i
                            update = 1
            if dist[t] == INF:
                return -1

            d = f; v = t
            while v != s:
                d = min(d, G[prevv[v]][preve[v]][1])
                v = prevv[v]
            f -= d
            res += d * dist[t]
            v = t
            while v != s:
                e = G[prevv[v]][preve[v]]
                e[1] -= d
                G[v][e[3]][1] += d
                v = prevv[v]
        return res

Verified

  • AOJ: "GRL_6_B: Network Flow - Minimum Cost Flow": source (Python3, 0.08sec)

with Dijkstra’s Algorithm

ポテンシャルを導入して、ダイクストラ法を用いて最短路を計算するようにしたもの。

計算量

\(O(F |E| \log |V|)\)

実装

from heapq import heappush, heappop
class MinCostFlow:
    INF = 10**18

    def __init__(self, N):
        self.N = N
        self.G = [[] for i in range(N)]

    def add_edge(self, fr, to, cap, cost):
        G = self.G
        G[fr].append([to, cap, cost, len(G[to])])
        G[to].append([fr, 0, -cost, len(G[fr])-1])

    def flow(self, s, t, f):
        N = self.N; G = self.G
        INF = MinCostFlow.INF

        res = 0
        H = [0]*N
        prv_v = [0]*N
        prv_e = [0]*N

        while f:
            dist = [INF]*N
            dist[s] = 0
            que = [(0, s)]

            while que:
                c, v = heappop(que)
                if dist[v] < c:
                    continue
                for i, (w, cap, cost, _) in enumerate(G[v]):
                    if cap > 0 and dist[w] > dist[v] + cost + H[v] - H[w]:
                        dist[w] = r = dist[v] + cost + H[v] - H[w]
                        prv_v[w] = v; prv_e[w] = i
                        heappush(que, (r, w))
            if dist[t] == INF:
                return -1

            for i in range(N):
                H[i] += dist[i]

            d = f; v = t
            while v != s:
                d = min(d, G[prv_v[v]][prv_e[v]][1])
                v = prv_v[v]
            f -= d
            res += d * H[t]
            v = t
            while v != s:
                e = G[prv_v[v]][prv_e[v]]
                e[1] -= d
                G[v][e[3]][1] += d
                v = prv_v[v]
        return res

Verified

  • AOJ: "GRL_6_B: Network Flow - Minimum Cost Flow": source (Python3, 0.04sec)



1. プログラミングコンテストチャレンジブック [第2版] p.199