PAST1L

L - Gradient

• Thoughts.
  • Connect 30 vertices at the lowest cost
  • There are five auxiliary vertices that may or may not be connected.
  • That’s nasty.
  • Find the minimum global tree with respect to “whether to pick 5 vertices” of 2^5?
    • Maybe we could start with the larger ones since removing the vertices won’t improve the cost, but we’ll get there in time without having to do that.
• Official Explanation - Steiner tree Issue. - But since T=30, Dreyfus-Wagner O(n 3^t + n^2 2^t + n^3) will not make it - minimum area tree - Kruskal method O(E log E)
  • 35 vertices and about 10^3 edges, so there’s plenty of time to search for all 2^5 patterns.
• 1WA python

def solve(N, M, LARGE, SMALL):
    from math import sqrt
    edges = []
    for i in range(N):
        x1, y1, c1 = LARGE[i]
        for j in range(i + 1, N):
            x2, y2, c2 = LARGE[j]
            d = sqrt((x1 - x2) ** 2 + (y1 - y2) ** 2)
            if c1 != c2:
                d *= 10
            edges.append((d, i, j))
    large_edges = edges[:]

    INF = 9223372036854775807
    ret = INF
    for subset in range(2 ** M):
        edges = large_edges[:]
        for m in range(M):
            if subset & (1 << m):
                x2, y2, c2 = SMALL[m]
                for i in range(N):
                    x1, y1, c1 = LARGE[i]
                    d = sqrt((x1 - x2) ** 2 + (y1 - y2) ** 2)
                    if c1 != c2:
                        d *= 10
                    edges.append((d, i, N + m))

        init_unionfind(N + M)
        edges.sort()
        cost = 0
        for d, i, j in edges:
            if is_connected(i, j):
                continue
            unite(i, j)
            cost += d
        ret = min(ret, cost)
    return ret

PAST1L

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