algorithm - 不能满足所有需求的最小成本的最大流量

Question

我正在使用 NetworkX求解具有多个源和汇的最大流问题。我发现了一个在 NetworkX 中运行得相对较好的函数，称为 max_cost_flow然而，我遇到的问题是它要求净需求为零，换句话说，任何汇都不应少于它的需求，否则会引发错误。

我可以使用什么(或如何修改此算法)来让它计算出可能的最佳流量，而不一定是满足所有条件的流量？

根据 kraskevich 的建议:

import networkx as nx

def convert(graph):

    allNodes = graph.nodes()

    newSource = len(allNodes) + 1
    newSink = len(allNodes) + 2

    graph.add_node(newSource)
    graph.add_node(newSink)


    for currentNode in allNodes:

        demand = graph.node[currentNode]['demand']

        if demand < 0 :
            graph.add_edge(newSource, currentNode, weight=0, capacity=demand)


        if demand > 0:
            graph.add_edge(newSink, currentNode, weight=0, capacity=demand)

    return graph



g = nx.DiGraph()

g.add_node(1, demand = 1)
g.add_node(2, demand = -2)
g.add_node(3, demand = 2)
g.add_node(4, demand = -4)

g.add_edge(1, 2, weight=4, capacity=100)
g.add_edge(1, 4, weight=3, capacity=100)
g.add_edge(3, 2, weight=5, capacity=100)
g.add_edge(3, 4, weight=2, capacity=100)
g.add_edge(3, 1, weight=1)
newGraph = convert(g)
print(nx.max_flow_min_cost(g, newGraph.nodes()[-2],newGraph.nodes()[-1]))

Answer 1

1. 您可以将多源流问题转换为单源问题，方法是创建一个新的源顶点并添加具有零成本的边以及从它到所有旧源的旧需求值。

2. 您可以对所有水槽做同样的事情(但边缘应该从旧水槽到新水槽)。

3. 之后，您可以使用 max_flow_min_cost函数以最小成本找到最大流量(不需要满足所有需求)。

更新:您的代码中存在一些错误。这是一个工作示例(我稍微更改了图表以使流量不为零):

import networkx as nx

def convert(graph):
    allNodes = graph.nodes()
    newSource = len(allNodes) + 1
    newSink = len(allNodes) + 2

    graph.add_node(newSource)
    graph.add_node(newSink)

    for currentNode in allNodes:
        demand = graph.node[currentNode]['demand']
        # Direction matters
        # It goes FROM the new source and TO the new sink
        if demand < 0:
            graph.add_edge(newSource, currentNode, weight=0, capacity=-demand)
        if demand > 0:
            graph.add_edge(currentNode, newSink, weight=0, capacity=demand)
        # We also need to zero out all demands
        graph.node[currentNode]['demand'] = 0
    return graph



g = nx.DiGraph()

g.add_node(1, demand = 1)
g.add_node(2, demand = -2)
g.add_node(3, demand = 2)
g.add_node(4, demand = -4)

g.add_edge(1, 2, weight=4, capacity=100)
g.add_edge(1, 4, weight=3, capacity=100)
g.add_edge(2, 3, weight=5, capacity=100)
g.add_edge(4, 3, weight=2, capacity=100)
g.add_edge(1, 3, weight=1)
newGraph = convert(g)
# The order of s and t matters here, too
print(nx.max_flow_min_cost(g, g.nodes()[-2], g.nodes()[-1]))