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$ curl repovive.com/roadmaps/graph-theory/mixed-practice-tree-fundamentals/max-white-subtree-implementation

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$ curl repovive.com/roadmaps/graph-theory/mixed-practice-tree-fundamentals/max-white-subtree-implementation

Repovive

Max White Subtree - Implementation - Mixed Practice: Tree Fundamentals | Graph Theory | Repovive

Repovive.

Graph Theory37 sections · 1633 units81h total

Open in Course

Max White Subtree - Implementation

(Rerooting with max)

Here is the solution:

c = array of node colors (1 or -1)
dp = array of size n
ans = array of size n

function dfs1(u, p):
    dp[u] = c[u]
    for v in adj[u]:
        if v == p:
            continue
        dfs1(v, u)
        dp[u] = dp[u] + max(0, dp[v])

function dfs2(u, p, up):
    ans[u] = dp[u] + max(0, up)
    for v in adj[u]:
        if v == p:
            continue
        newUp = c[u] + max(0, up) + (dp[u] - c[u] - max(0, dp[v]))
        dfs2(v, u, newUp)

This runs in $O(n)$ time and uses $O(n)$ space.