用于检查图形是否为非循环的递归函数

Question

我在python中使用递归函数有点困惑.Below是我在python中的代码，它检查给定的一组点是否形成一个acylcic图。输入格式如下：输入中的前两个数字由顶点数和边数组成。下面的剩余数字表示形成边缘的顶点，其方向是从左数到右数。例如，检查下面的图像

以下是解决此问题的代码。输出提供图中的循环次数

#Uses python3

import sys

def acyclic(adj,cc,in_path):
    # check if value of ad
    global clock
    global cycles
    clock += 1
    for i in adj:
        for j in i:
            if visited[j] != 1:
                num = reach(adj,j,cc,in_path)

                if (0 not in visited):
                    return cc
                cc += 1

    return cc

def reach(adj,x,cc,in_path):
    visited[x] = 1
    # preVisit(x)
    global cycles
    cnum[x] = cc
    print(x)
    in_path.append(x)
    print(in_path)
    for i in adj[x]:
        print(adj[x])
        if visited[i] != 1:
            return reach(adj,i,cc,in_path)
        elif i in in_path:
            # print("in path")
            cycles += 1
            return (cycles)
            break
    # postVisit(x)
    # return visited[x]

def dfs(adj):
    global m
    global cycles
    # print(m)
    for i in range(n):
        in_path = []
        if visited[i] != 1:
            reach(adj,i,0,in_path)
    print (cycles)



def preVisit(x):
    global clock
    clock += 1
    prev[x] = clock
    # print("prev")
    # print(prev)


def postVisit(x):
    global clock
    clock += 1
    postV[x] = clock
    # print("post")
    # print(postV)




if __name__ == '__main__':
    cycles = 0
    clock = 0
    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]))
    prev = [0 for _ in range(n)]
    postV = [0 for _ in range(n)]
    adj = [[] for _ in range(n)]
    cnum = [0 for i in range(n)]
    visited = [0 for i in range(n)]
    for (a, b) in edges:
        adj[a - 1].append(b - 1)
    print(adj)
    dfs(adj)

每当我在函数return "reach(adj,i,cc,in_path)"中使用"return(cycles)"和reach()时，我都会得到正确答案。但是，取下它们会给我错误的答案。有人可以解释一下为什么会这样吗？

提前致谢

此致

Rishit