任何人都可以向我解釋 Kosaraju 尋找連通分量的算法背后的邏輯嗎？我已閱讀說明，但我無法理解反向圖上的 DFS 如何檢測強連接組件的數(shù)量。def dfs(visited, stack, adj, x):    visited[x] = 1    for neighbor in adj[x]:        if (visited[neighbor]==0):            dfs(visited, stack, adj, neighbor)    stack.insert(0, x)    return stack, visiteddef reverse_dfs(visited, adj, x, cc):    visited[x] = 1    for neighbor in adj[x]:        if (visited[neighbor]==0):            cc += 1            reverse_dfs(visited, adj, neighbor,cc)    print(x)    return ccdef reverse_graph(adj):    vertex_num = len(adj)    new_adj = [ [] for _ in range(vertex_num)]    for i in range(vertex_num):        for j in adj[i]:            new_adj[j].append(i)    return new_adjdef find_post_order(adj):    vertex_num = len(adj)    visited = [0] * vertex_num    stack = []    for vertex in range(vertex_num):        if visited[vertex] == 0:            stack, visited = dfs(visited, stack, adj, vertex)    return stackdef number_of_strongly_connected_components(adj):    vertex_num = len(adj)    new_adj = reverse_graph(adj)    stack = find_post_order(adj)    visited = [0] * vertex_num    cc_num = 0    while (stack):        vertex = stack.pop(0)        print(vertex)        print('------')        if visited[vertex] == 0:            cc_num = reverse_dfs(visited, new_adj, vertex, cc_num)    return cc_numif __name__ == '__main__':    input = sys.stdin.read()    data = list(map(int, input.split()))    n, m = data[0:2]    data = data[2:]    edges = list(zip(data[0:(2 * m):2], data[1:(2 * m):2]))    adj = [[] for _ in range(n)]    for (a, b) in edges:        adj[a - 1].append(b - 1)    print(number_of_strongly_connected_components(adj))在上面你可以找到我的實現(xiàn)。我猜初始 DFS 和反向操作已正確完成，但我無法正確執(zhí)行第二個 DFS。提前致謝。