28 lines
834 B
Python
28 lines
834 B
Python
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"""
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simple union find with path compression:
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at first, each vertices are disjoint, there are N connected components at the begining,
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after add each edge, we unify those two connected components, if two vertices are already
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in the same connected component, then this edge will form a circle, just return this edge
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"""
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def findRedundantConnection(self, edges):
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id = list(range(len(edges) + 1))
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def find(u):
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root = u
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while root != id[root]: root = id[root]
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# path compression:
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while u != root:
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next = id[u]
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id[u] = root
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u = next
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return root
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for u, v in edges:
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root1, root2 = find(u), find(v)
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if root1 == root2: return [u, v]
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# else, unify these two components:
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id[root1] = root2
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