def min_bridge_cost(n, costs):

    costs.sort(key=lambda x: x[2])

    parent = [i for i in range(n)]

    def find(x):
        if parent[x] != x:
            parent[x] = find(parent[x])
        return parent[x]

    def union(a, b):
        a = find(a)
        b = find(b)
        if a != b:
            parent[b] = a
            return True
        return False

    total = 0

    for a, b, cost in costs:
        if union(a, b):
            total += cost

    return total


n = 4
costs = [[0,1,1], [0,2,2], [1,2,5], [1,3,1], [2,3,8]]

print(min_bridge_cost(n, costs))