时间复杂度 $O(nV)$

import java.io.*;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        solve();
    }

    static int size[];
    static List<Integer> list;

    private static void solve() {
        int n = IN.nextInt(), V = IN.nextInt(), vs[] = new int[n + 1], ws[] = new int[n + 1];
        init(n);
        int fa, root = -1;
        for (int i = 1; i <= n; i++) {
            vs[i] = IN.nextInt();
            ws[i] = IN.nextInt();
            if ((fa = IN.nextInt()) == -1) {
                root = i;
            } else {
                add(fa, i);
            }
        }
        dfs(root);
        int dp[][] = new int[n + 1][V + 1], cur;
        for (int i = 1; i <= n; i++) {
            cur = list.get(i - 1);
            for (int j = V; j >= 0; j--) {
                // 若不选物品i，则也不可以选它的子树
                dp[i][j] = dp[i - size[cur]][j];
                if (j >= vs[cur]) {
                    // 若选了物品i，则可以选它的子树
                    dp[i][j] = Math.max(dp[i][j], dp[i - 1][j - vs[cur]] + ws[cur]);
                }
            }
        }
        System.out.println(dp[n][V]);
    }

    public static void dfs(int cur) {
        size[cur] = 1;
        for (int i = head[cur]; i != 0; i = next[i]) {
            dfs(to[i]);
            size[cur] += size[to[i]];
        }
        list.add(cur);
    }

    static int head[], next[], to[], id = 1;

    public static void init(int n) {
        head = new int[n + 2];
        next = new int[n + 2];
        to = new int[n + 2];

        list = new ArrayList<>(n + 1);
        size = new int[n + 1];
    }

    public static void add(int f, int t) {
        next[id] = head[f];
        head[f] = id;
        to[id++] = t;
    }

    static class IN {
        private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in), 65535);
        private static StringTokenizer st = null;
        public static int nextInt() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {
                    break;
                }
            }
            return Integer.valueOf(st.nextToken());
        }
    }
}