拓扑排序python
我编写了一个DFS非递归解决方案，但我无法对其进行修改以进行拓扑排序：
def dfs(graph,start):
path = []
stack = [start]
while stack != []:
v = stack.pop()
if v not in path: path.append(v)
for w in reversed(graph[v]):
if w not in path and not w in stack:
stack.append(w)
return path
任何想法如何修改它？
随着递归版本，我可以很容易地进行排序：
def dfs_rec(graph,start,path):
path = path + [start]
for edge in graph[start]:
if edge not in path:
path = dfs_rec(graph, edge,path)
print start
return path
输入：
>>> graph = {
1: [2, 3],
2: [4, 5, 6],
3: [4,6],
4: [5,6],
5: [6],
6: []
}
>>> dfs_rec(graph,1,[])
6
5
4
2
3
1
[1, 2, 4, 5, 6, 3]
>>> dfs(graph,1)
[1, 2, 4, 5, 6, 3]
>>> graph = {
1: [3],
3: [5,6],
5: [4],
4: [7],
7: [],
6: []
}
>>> print dfs_rec(graph,1,[])
7
4
5
6
3
1
[1, 3, 5, 4, 7, 6]
>>> print dfs(graph,1)
[1, 3, 5, 4, 7, 6]
所以我需要在非递归中得到这个顺序。
非递归解决方案：
我认为这也可能是解决方案，如果我错了，请告诉我。
def dfs(graph,start):
path = []
stack = [start]
label = len(graph)
result = {}
while stack != []:
#this for loop could be done in other ways also
for element in stack:
if element not in result:
result[element] = label
label = label - 1
v = stack.pop()
if v not in path: path.append(v)
for w in reversed(graph[v]):
if w not in path and not w in stack:
stack.append(w)
result = {v:k for k, v in result.items()}
return path,result
输入：
graph = { 1: [3], 3:[5,6] , 5:[4] , 4:[7], 7:[],6:[]}
print dfs(graph,1)
输出：
([1, 3, 5, 4, 7, 6], {1: 7, 2: 4, 3: 5, 4: 6, 5: 3, 6: 1})
1
/
3
/
5 6
/
4
/
7
FWIW，这是我为非递归拓扑排序所做的一些代码。
from collections import defaultdict, namedtuple
from itertools import islice
Results = namedtuple('Results', ['sorted', 'cyclic'])
def topological_sort(dependency_pairs):
'Sort values subject to dependency constraints'
num_heads = defaultdict(int) # num arrows pointing in
tails = defaultdict(list) # list of arrows going out
heads = [] # unique list of heads in order first seen
for h, t in dependency_pairs:
num_heads[t] += 1
if h in tails:
tails[h].append(t)
else:
tails[h] = [t]
heads.append(h)
ordered = [h for h in heads if h not in num_heads]
for h in ordered:
for t in tails[h]:
num_heads[t] -= 1
if not num_heads[t]:
ordered.append(t)
cyclic = [n for n, heads in num_heads.items() if heads]
return Results(ordered, cyclic)
if __name__ == '__main__':
print( topological_sort('aa'.split()) )
print( topological_sort('ah bg cf ch di ed fb fg hd he ib'.split()) )
#this algorithm gives the logic of topological sorting..if u want to run this
#give adjacency mat of your choice and this algorithm works on graph elements ranging from 0 to n
a=[[0,0,1,0,0,0],[0,0,1,0,0,0],[0,0,0,1,1,0],[0,0,0,0,1,0],[0,0,0,0,0,0],[0,0,1,0,0,0]]
vis=[0 for i in range(0,len(a))]
s=[]
orderstack=[]#stores the reverse order of topological sorted elements
def dfs_for_topological_sorting(a,vis,i):
vis[i]=1
x=0
for j in range(0,len(a[0])):
if(a[i][j]==1 and vis[j]==0):
x=1
s.append(j)
#print(s)
dfs_for_topological_sorting(a,vis,j)
if(x==0 and len(s)!=0):
orderstack.append(s[len(s)-1])
if(len(s)>0):
dfs_for_topological_sorting(a,vis,s.pop())
for i in range(0,len(a)):
if(i not in orderstack):
s.append(i)
dfs_for_topological_sorting(a,vis,i)
print(orderstack[len(orderstack)-1::-1])