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Python 判断 有向图 是否有环的实例讲解

实例如下:

import numpy
from numpy import *
def dfs( v ):
 vis[v] = -1
 flag = 0
 for i in range(n):
 # print (a[v][i],'---', vis[i] )
 if a[v][i] != 0 and vis[i] != -1:
  dfs(i)
  vis[i] = 1
 else:
  pass
 if a[v][i] != 0 and vis[i] == -1:
  print ('Yes, there is A loop in this network\n')
  global swi
  swi = True
  exit()
  return
  # break
 else:
  pass
 print ('s = 0')
 return False

global swi
swi = False
'''===装载数据'''
edges = numpy.loadtxt('9_nodes_with_r_edge_8_to_3.txt')

# edges = [ int(i) for i in edges]
bian = int(shape(edges)[0]) - 1
print (bian,'edges in the network \n')
print (shape(edges),'\n')

n = int( edges[0][1] )

c = int( edges[0][0] )
# n, c = input().split()
# n = int(n)
# c = int(c)
a = [([0] * n) for i in range(n)]
vis = [0] * c
for i in range(1, c+1):
 s, t = edges[i][0:2]
 # print (s,' - ', t )
 '''GO_OBO文件则 s, t 不需要 -1 '''
 s = int(s) - 1
 t = int(t) - 1
 # s = int(s)
 # t = int(t)
 a[s][t] = 1
# print (a)
# print (vis)
dfs(0)
# print (swi)
if not swi:
 print('No loop, DAG - DAG - DAG')

用到 numpy 模块,读取的 txt 文件为 有向图的连边,其中第一行 第一个数字 为 边的数量,第二个数字为 节点数 第二行及以后 前两个数字,第一个为 起点, 第二个为 落点。

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