python tarjan算法实现

#coding:utf-8
#tarjan 算法
#https://blog.csdn.net/jeryjeryjery/article/details/52829142?locationNum=4&fps=1
#求任意顶点开始的联通图 有且仅存在一个 且dfn[u] == low[u]
from collections import OrderedDict
matric = [[0,1,1,0,0,0],[0,0,0,1,0,0],[0,0,0,1,1,0],[1,0,0,0,0,1],[0,0,0,0,0,1],[0,0,0,0,0,0]]
dfn = OrderedDict()
low = OrderedDict()
flag = dict()
count = 0
n = 6 
num = 0
class Stack(object):
    def __init__(self):
        self.items = list()
    def push(self, item):
        self.items.append(item)
    def pop(self):
        return self.items.pop()
    def clear(self):
        del self.items[:]
    def empty(self):
        return self.size() == 0
    def size(self):
        return len(self.items)
    def top(self):
        return self.items[self.size() - 1]

s = Stack()
def tarjan(u):
    global s,num,n,count,flag,stack,dfn,low,matric
    count = count + 1
    dfn[u] = low[u] = count
    s.push(u)
    flag[u] = True
    #print("visiting {0} ...".format(str(u + 1)))
    for i in range(n):
        if matric[u][i] == 0:
            continue
        if flag.get(i, False) is True:
            if (dfn[i] < low[u]):
                low[u] = dfn[i]
        else:
            tarjan(i)
            low[u] = min(low[u], low[i])
    if (dfn[u] == low[u] and s.empty() is False):
       print("********连通图********")
       m = s.pop()
       flag[m] = False
       print(m + 1)
       while m != u and s.empty() is False:
           num = num + 1
           m = s.pop()
           flag[m] = False
           print(m+1)
       print("*********************")  

if __name__  == "__main__":
    pass
    tarjan(3)
    print("连通图数量...")
    print(num)

tarjan 算法 求任意顶点开始的 连通图 （相对于一般算法 每个顶点 只进行一次正向深度优先搜索）