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# 3108. Minimum Cost Walk in Weighted Graph
"""
there is an undirected weighted graph with 'n' vertices labeled from 0 to 'n
- 1'. you are given the integer 'n' and an array 'edges', where 'edges[i] =
[u_i, v_i, w_i]' indicates that there is an edge between vertices 'u_i' and
'v_i' with a weight of 'w_i'. a walk on the graph is a sequence of vertices
and edges. the walk starts and ends with a vertex, and each edge connects the
vertex that comes before it and the vertex that comes after it. it's
important to note that a walk may visit the same edge or vertex more than
once. return the array 'answer' where 'answer[i]' denotes the minimum cost of
a walk for query 'i'.
"""
class UnionFind:
def __init__(self, size):
self.parent = [i for i in range(size)]
self.rank = [0 for _ in range(size)]
self.d = [~0 for _ in range(size)]
def find(self, node):
if self.parent[node] != node:
self.parent[node] = self.find(self.parent[node])
return self.parent[node]
def union(self, a, b, val):
p1, p2 = self.find(a), self.find(b)
if self.rank[p1] > self.rank[p2]:
self.parent[p2] = p1
self.d[p1] &= self.d[p2] & val
elif self.rank[p1] < self.rank[p2]:
self.parent[p1] = p2
self.d[p2] &= self.d[p1] & val
else:
self.parent[p1] = p2
self.d[p2] &= self.d[p1] & val
self.rank[p2] += 1
class Solution(object):
def minimumCost(self, n, edges, query):
"""
:type n: int
:type edges: List[List[int]]
:type query: List[List[int]]
:rtype: List[int]
"""
uf = UnionFind(n)
for u, v, w in edges:
uf.union(u, v, w)
ans = []
for u, v in query:
if uf.find(u) != uf.find(v):
ans.append(-1)
else:
ans.append(uf.d[uf.find(u)])
return ans
if __name__ == "__main__":
obj = Solution()
print(
obj.minimumCost(
n=5, edges=[[0, 1, 7], [1, 3, 7], [1, 2, 1]], query=[[0, 3], [3, 4]]
)
)
print(
obj.minimumCost(
n=3, edges=[[0, 2, 7], [0, 1, 15], [1, 2, 6], [1, 2, 1]], query=[[1, 2]]
)
)
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