📅  最后修改于: 2023-12-03 15:22:09.662000             🧑  作者: Mango
Bellman Ford 算法可以用来在带负边权的图中找到最短路径,它的时间复杂度为$O(VE)$。在最小成本最大流问题中,我们需要在残余网络中找到从源点到汇点的一条路径,该路径上的边权和为负,并且路径上的最小流量即为该路径的最大流量。
下面是使用 Bellman Ford 算法从图中得到的最小成本最大流的 Python 代码实现:
def min_cost_max_flow(graph, source, sink):
# 初始化残余流量矩阵和残余容量矩阵
flow = [[0 for _ in range(len(graph))] for _ in range(len(graph))]
capacity = [[graph[i][j] for j in range(len(graph))] for i in range(len(graph))]
# 初始化距离矩阵
distance = [float('inf') for _ in range(len(graph))]
distance[source] = 0
# 执行 Bellman Ford 算法,找到最短路径
for i in range(len(graph) - 1):
for j in range(len(graph)):
for k in range(len(graph)):
if capacity[j][k] > 0 and distance[j] + graph[j][k] < distance[k]:
distance[k] = distance[j] + graph[j][k]
# 寻找一条从源点到汇点的路径,该路径上的边权和为负
while True:
# 使用 Dijkstra 算法找到最短路径
parent = [-1 for _ in range(len(graph))]
distance = [float('inf') for _ in range(len(graph))]
distance[source] = 0
queue = [source]
while queue:
u = queue.pop(0)
for v in range(len(graph)):
if capacity[u][v] > 0 and distance[u] + graph[u][v] < distance[v]:
distance[v] = distance[u] + graph[u][v]
parent[v] = u
queue.append(v)
# 如果找不到路径,则退出
if parent[sink] == -1:
break
# 计算路径上的最小流量
min_flow = float('inf')
v = sink
while v != source:
u = parent[v]
min_flow = min(min_flow, capacity[u][v] - flow[u][v])
v = u
# 更新流和容量矩阵
v = sink
while v != source:
u = parent[v]
flow[u][v] += min_flow
flow[v][u] -= min_flow
capacity[u][v] -= min_flow
capacity[v][u] += min_flow
v = u
# 计算最小成本和最大流量
min_cost = 0
max_flow = 0
for i in range(len(graph)):
for j in range(len(graph)):
if flow[i][j] > 0:
max_flow += flow[i][j]
min_cost += flow[i][j] * graph[i][j]
return min_cost, max_flow
以上代码实现中,首先初始化残余流量矩阵和残余容量矩阵,并将距离矩阵初始化为无穷大。然后执行 Bellman Ford 算法找到最短路径,如果该路径上的边权和为负,则使用 Dijkstra 算法找到该路径上的最小流量,并更新流和容量矩阵。最后计算最小成本和最大流量并返回。