费用流模型 -- 关于 pywrapgraph 的介绍
https://developers.google.com/optimization/flow/mincostflow
对于一个n个点，m条边的网络，
第第i条边，它的方向是 start_nodes[i] -> end_nodes[i], 这条边的容量为capacities[i]，费用为unit_costs[i]
对第i个点，当supplies[i]为正，表示他为供给点(源点)；当supplies[i]为负，表示他为需求点(汇点)
例子：
# Define four parallel arrays: start_nodes, end_nodes, capacities, and unit costs
# between each pair. For instance, the arc from node 0 to node 1 has a
# capacity of 15 and a unit cost of 4.
start_nodes = [ 0, 0, 1, 1, 1, 2, 2, 3, 4]
end_nodes = [ 1, 2, 2, 3, 4, 3, 4, 4, 2]
capacities = [15, 8, 20, 4, 10, 15, 4, 20, 5]
unit_costs = [ 4, 4, 2, 2, 6, 1, 3, 2, 3]
# Define an array of supplies at each node.
supplies = [20, 0, 0, -5, -15]
初始化一个最小费用流的类(解决者)
min_cost_flow = pywrapgraph.SimpleMinCostFlow()
添加一条带有容量和费用的边
for i in range(0, len(start_nodes)):
min_cost_flow.AddArcWithCapacityAndUnitCost(start_nodes[i], end_nodes[i],
capacities[i], unit_costs[i])
为每个点设置supply, 如果为正，表示为源点；如果为负，表示为汇点
for i in range(0, len(supplies)):
min_cost_flow.SetNodeSupply(i, supplies[i])
SolveMaxFlowWithMinCost
返回值类型：status
计算最小费用最大流，最大流将由函数MaximumFlow()给出
OptimalCost() 得到最小费用
MaximumFlow() 得到最大流
NumArcs() 返回网络中的边数
Flow() 返回那条边有多少流量