Question

我有这个问题，我被困在了。

给定网络N，找到最小切割次数所需的时间复杂度：Poly（| N |）*＃（min cut）。

我没有成功找到任何有用的东西，只是如何通过在残差图中从S开始使用BFS来找到第一个最小切割。

感谢。

Answer 1

在最坏的情况下，最小s−t削减复杂度的数量可以是指数级的。很容易构建具有独特最小切割的流动网络。

# Python program for finding min-cut in the given graph
# Complexity : (E*(V^3))     
# This class represents a directed graph using adjacency matrix 
class Graph:

    def __init__(self,graph):
        self.graph = graph # residual graph
        self.org_graph = [i[:] for i in graph]
        self. ROW = len(graph)
        self.COL = len(graph[0])


    '''Returns true if there is a path from source 's' to sink 't' in
    residual graph. Also fills parent[] to store the path '''
    def BFS(self,s, t, parent):

        # Mark all the vertices as not visited
        visited =[False]*(self.ROW)

        # Create a queue for BFS
        queue=[]

        # Mark the source node as visited and enqueue it
        queue.append(s)
        visited[s] = True

         # Standard BFS Loop
        while queue:

            #Dequeue a vertex from queue and print it
            u = queue.pop(0)

            # Get all adjacent vertices of the dequeued vertex u
            # If a adjacent has not been visited, then mark it
            # visited and enqueue it
            for ind, val in enumerate(self.graph[u]):
                if visited[ind] == False and val > 0 :
                    queue.append(ind)
                    visited[ind] = True
                    parent[ind] = u

        # If we reached sink in BFS starting from source, then return
        # true, else false
        return True if visited[t] else False


    # Returns tne min-cut of the given graph
    def minCut(self, source, sink):

        # This array is filled by BFS and to store path
        parent = [-1]*(self.ROW)

        max_flow = 0 # There is no flow initially

        # Augment the flow while there is path from source to sink
        while self.BFS(source, sink, parent) :

            # Find minimum residual capacity of the edges along the
            # path filled by BFS. Or we can say find the maximum flow
            # through the path found.
            path_flow = float("Inf")
            s = sink
            while(s !=  source):
                path_flow = min (path_flow, self.graph[parent[s]][s])
                s = parent[s]

            # Add path flow to overall flow
            max_flow +=  path_flow

            # update residual capacities of the edges and reverse edges
            # along the path
            v = sink
            while(v !=  source):
                u = parent[v]
                self.graph[u][v] -= path_flow
                self.graph[v][u] += path_flow
                v = parent[v]

        # print the edges which initially had weights
        # but now have 0 weight
        for i in range(self.ROW):
            for j in range(self.COL):
                if self.graph[i][j] == 0 and self.org_graph[i][j] > 0:
                    print str(i) + " - " + str(j)


# Create a graph given in the above diagram
graph = [[0, 16, 13, 0, 0, 0],
        [0, 0, 10, 12, 0, 0],
        [0, 4, 0, 0, 14, 0],
        [0, 0, 9, 0, 0, 20],
        [0, 0, 0, 7, 0, 4],
        [0, 0, 0, 0, 0, 0]]

g = Graph(graph)

source = 0; sink = 5

g.minCut(source, sink)

最大流量的最小切割次数

1 个答案: