动态规划(三) —— 股票投资系列问题总结

**动态规划(三) —— 股票投资系列问题总结**

在前两篇文章中，我们已经介绍了动态规划的基本概念、算法设计思想以及几个经典的问题。今天，我们将继续讨论股票投资系列问题，总结这些问题的解决方案和代码实现。

**1. 最小成本流**

最小成本流是指从源点到汇点的最短路径，其权重为每条边的成本。这个问题可以使用动态规划来解决。

def min_cost_flow(graph, source, sink):
 # 初始化距离和前驱节点 distance = [float('inf')] * len(graph)
 predecessor = [-1] * len(graph)

 # 初始化队列 queue = [(0, source)]

 while queue:
 d, u = heapq.heappop(queue)

 if distance[u] < d:
 continue for v in graph[u]:
 cost = graph[u][v]
 new_distance = distance[u] + cost if new_distance < distance[v]:
 distance[v] = new_distance predecessor[v] = u heapq.heappush(queue, (new_distance, v))

 # 回溯路径并计算成本 path_cost =0 current_node = sink while current_node != source:
 previous_node = predecessor[current_node]
 cost = graph[previous_node][current_node]

 path_cost += cost current_node = previous_node return distance[sink], path_cost

def max_flow(graph, source, sink):
 # 初始化距离和前驱节点 distance = [float('inf')] * len(graph)
 predecessor = [-1] * len(graph)

 # 初始化队列 queue = [(0, source)]

 while queue:
 d, u = heapq.heappop(queue)

 if distance[u] < d:
 continue for v in graph[u]:
 cost = graph[u][v]
 new_distance = distance[u] + cost if new_distance < distance[v]:
 distance[v] = new_distance predecessor[v] = u heapq.heappush(queue, (new_distance, v))

 # 回溯路径并计算流量 flow =0 current_node = sink while current_node != source:
 previous_node = predecessor[current_node]
 capacity = graph[previous_node][current_node]

 if capacity >0:
 flow += capacity current_node = previous_node return flow

def stock_investment(prices, days):
 # 初始化距离和前驱节点 distance = [float('inf')] * (days +1)
 predecessor = [-1] * (days +1)

 # 初始化队列 queue = [(0,0)]

 while queue:
 d, u = heapq.heappop(queue)

 if distance[u] < d:
 continue for v in range(u +1, days +1):
 cost = prices[v -1]
 new_distance = distance[u] + cost if new_distance < distance[v]:
 distance[v] = new_distance predecessor[v] = u heapq.heappush(queue, (new_distance, v))

 # 回溯路径并计算收益 profit =0 current_day = days while current_day !=0:
 previous_day = predecessor[current_day]
 cost = prices[previous_day]

 profit += cost current_day = previous_day return profit