// Assume each node has: vector<Node*> children and an integer cost (or weight) for removal.
int dp(Node* u) {
    // Base case: if u is a leaf, return its removal cost.
    if(u->children.empty()) {
        return u->cost;
    }
    // Otherwise, compute the cost if we do not remove u:
    int sumChildren = 0;
    for(Node* v : u->children) {
        sumChildren += dp(v);
    }
    // The cost for u is the minimum of:
    // 1. Removing u directly (which cuts off its whole subtree)
    // 2. Not removing u and instead cutting at the children.
    return min(u->cost, sumChildren);
}

// To get the answer, since the root cannot be cut:
int answer(Node* root) {
    int total = 0;
    for(Node* child : root->children) {
        total += dp(child);
    }
    return total;
}