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01601e6382
* Add disjoint set * disjoint set: add doctest, make code more Pythonic * disjoint set: replace x.p with x.parent * disjoint set: add test and refercence
80 lines
1.7 KiB
Python
80 lines
1.7 KiB
Python
"""
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disjoint set
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Reference: https://en.wikipedia.org/wiki/Disjoint-set_data_structure
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"""
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class Node:
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def __init__(self, data):
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self.data = data
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def make_set(x):
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"""
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make x as a set.
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"""
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# rank is the distance from x to its' parent
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# root's rank is 0
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x.rank = 0
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x.parent = x
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def union_set(x, y):
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"""
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union two sets.
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set with bigger rank should be parent, so that the
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disjoint set tree will be more flat.
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"""
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x, y = find_set(x), find_set(y)
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if x.rank > y.rank:
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y.parent = x
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else:
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x.parent = y
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if x.rank == y.rank:
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y.rank += 1
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def find_set(x):
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"""
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return the parent of x
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"""
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if x != x.parent:
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x.parent = find_set(x.parent)
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return x.parent
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def find_python_set(node: Node) -> set:
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"""
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Return a Python Standard Library set that contains i.
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"""
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sets = ({0, 1, 2}, {3, 4, 5})
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for s in sets:
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if node.data in s:
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return s
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raise ValueError(f"{node.data} is not in {sets}")
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def test_disjoint_set():
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"""
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>>> test_disjoint_set()
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"""
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vertex = [Node(i) for i in range(6)]
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for v in vertex:
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make_set(v)
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union_set(vertex[0], vertex[1])
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union_set(vertex[1], vertex[2])
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union_set(vertex[3], vertex[4])
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union_set(vertex[3], vertex[5])
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for node0 in vertex:
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for node1 in vertex:
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if find_python_set(node0).isdisjoint(find_python_set(node1)):
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assert find_set(node0) != find_set(node1)
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else:
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assert find_set(node0) == find_set(node1)
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if __name__ == "__main__":
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test_disjoint_set()
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