-
Notifications
You must be signed in to change notification settings - Fork 312
Commit
This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository.
- Loading branch information
1 parent
7f54a87
commit d55d809
Showing
2 changed files
with
158 additions
and
0 deletions.
There are no files selected for viewing
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -0,0 +1,52 @@ | ||
""" | ||
Given a directed graph, find_SCC returns a list of lists containing | ||
the strongly connected components in topological order. | ||
Note that this implementation can be also be used to check if a directed graph is a | ||
DAG, and in that case it can be used to find the topological ordering of the nodes. | ||
""" | ||
|
||
def find_SCC(graph): | ||
SCC, S, P = [], [], [] | ||
depth = [0] * len(graph) | ||
|
||
stack = list(range(len(graph))) | ||
while stack: | ||
node = stack.pop() | ||
if node < 0: | ||
d = depth[~node] - 1 | ||
if P[-1] > d: | ||
SCC.append(S[d:]) | ||
del S[d:], P[-1] | ||
for node in SCC[-1]: | ||
depth[node] = -1 | ||
elif depth[node] > 0: | ||
while P[-1] > depth[node]: | ||
P.pop() | ||
elif depth[node] == 0: | ||
S.append(node) | ||
P.append(len(S)) | ||
depth[node] = len(S) | ||
stack.append(~node) | ||
stack += graph[node] | ||
return SCC[::-1] | ||
|
||
""" | ||
Given an undirected simple graph, find_BCC returns a list of lists | ||
containing the biconnected components of the graph. Runs in O(n + m) time. | ||
""" | ||
def find_BCC(graph): | ||
d = 0 | ||
depth = [0] * len(graph) | ||
stack = list(range(len(graph))) | ||
while stack: | ||
node = stack.pop() | ||
if node < 0: | ||
d -= 1 | ||
elif not depth[node]: | ||
d = depth[node] = d + 1 | ||
stack.append(~node) | ||
stack += graph[node] | ||
|
||
graph2 = [[v for v in g if d + 2 > depth[v] != d - 1] for g,d in zip(graph, depth)] | ||
return find_SCC(graph2) |
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -0,0 +1,106 @@ | ||
""" | ||
Binary lifting technique applied to a tree. | ||
There are three different uses of this implementation | ||
1. Computing LCA in O(log n) time. | ||
Example: | ||
0 | ||
| | ||
1 | ||
/ \ | ||
2 3 | ||
graph = [[1], [0, 2, 3], [1], [1]] | ||
BL = binary_lift(graph, root=0) | ||
print(BL.lca(2, 3)) # prints 1 | ||
2. Compute the distance between two nodes in O(log n) time. | ||
Example: | ||
graph = [[1], [0, 2, 3], [1], [1]] | ||
BL = binary_lift(graph) | ||
print(BL.distance(2, 3)) # prints 2 | ||
3. Compute the sum/min/max/... of the weight | ||
of a path between a pair of nodes in O(log n) time. | ||
res = Path[0] | ||
for node in Path[1:]: | ||
res = f(res, node) | ||
return res | ||
Example: | ||
graph = [[1], [0, 2, 3], [1], [1]] | ||
data = [1, 10, 20, 5] | ||
BL = binary_lift(graph, data, f = lambda a,b: a + b) | ||
print(BL(2, 3)) # prints 35 | ||
""" | ||
|
||
class binary_lift: | ||
def __init__(self, graph, data=(), f=min, root=0): | ||
n = len(graph) | ||
|
||
parent = [-1] * (n + 1) | ||
depth = self.depth = [-1] * n | ||
bfs = [root] | ||
depth[root] = 0 | ||
for node in bfs: | ||
for nei in graph[node]: | ||
if depth[nei] == -1: | ||
parent[nei] = node | ||
depth[nei] = depth[node] + 1 | ||
bfs.append(nei) | ||
|
||
data = self.data = [data] | ||
parent = self.parent = [parent] | ||
self.f = f | ||
|
||
for _ in range(max(depth).bit_length()): | ||
old_data = data[-1] | ||
old_parent = parent[-1] | ||
|
||
data.append([f(val, old_data[p]) for val,p in zip(old_data, old_parent)]) | ||
parent.append([old_parent[p] for p in old_parent]) | ||
|
||
def lca(self, a, b): | ||
depth = self.depth | ||
parent = self.parent | ||
|
||
if depth[a] < depth[b]: | ||
a,b = b,a | ||
|
||
d = depth[a] - depth[b] | ||
for i in range(d.bit_length()): | ||
if (d >> i) & 1: | ||
a = parent[i][a] | ||
|
||
for i in range(depth[a].bit_length())[::-1]: | ||
if parent[i][a] != parent[i][b]: | ||
a = parent[i][a] | ||
b = parent[i][b] | ||
|
||
if a != b: | ||
return parent[0][a] | ||
else: | ||
return a | ||
|
||
def distance(self, a, b): | ||
return self.depth[a] + self.depth[b] - 2 * self.depth[self.lca(a,b)] | ||
|
||
def __call__(self, a, b): | ||
depth = self.depth | ||
parent = self.parent | ||
data = self.data | ||
f = self.f | ||
|
||
c = self.lca(a, b) | ||
val = data[0][c] | ||
for x,d in (a, depth[a] - depth[c]), (b, depth[b] - depth[c]): | ||
for i in range(d.bit_length()): | ||
if (d >> i) & 1: | ||
val = f(val, data[i][x]) | ||
x = parent[i][x] | ||
|
||
return val |