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| 1 | +/** |
| 2 | + * @param {number} n |
| 3 | + * @param {number[]} present |
| 4 | + * @param {number[]} future |
| 5 | + * @param {number[][]} hierarchy |
| 6 | + * @param {number} budget |
| 7 | + * @return {number} |
| 8 | + */ |
| 9 | +var maxProfit = function (n, present, future, hierarchy, budget) { |
| 10 | + const freq = {} |
| 11 | + for (const [p, c] of hierarchy) { |
| 12 | + if (!freq[p]) freq[p] = [] |
| 13 | + freq[p].push(c) |
| 14 | + } |
| 15 | + const MINI = -(10 ** 9) |
| 16 | + const ans = dfs(1, freq, budget, present, future, MINI)[0] |
| 17 | + return Math.max(...ans) |
| 18 | + |
| 19 | + function dfs(u, freq, budget, present, future, MINI) { |
| 20 | + const dp = (freq[u] || []).map((child) => |
| 21 | + dfs(child, freq, budget, present, future, MINI), |
| 22 | + ) |
| 23 | + const a = Array(budget + 1).fill(MINI) |
| 24 | + const b = Array(budget + 1).fill(MINI) |
| 25 | + |
| 26 | + for (let v = 0; v < 2; v++) { |
| 27 | + let x = Array(budget + 1).fill(MINI) |
| 28 | + let y = Array(budget + 1).fill(MINI) |
| 29 | + x[0] = 0 |
| 30 | + const c = v === 0 ? present[u - 1] : Math.floor(present[u - 1] / 2) |
| 31 | + const profit = future[u - 1] - c |
| 32 | + if (c <= budget) { |
| 33 | + y[c] = profit |
| 34 | + } |
| 35 | + for (const [c0, c1] of dp) { |
| 36 | + const x1 = Array(budget + 1).fill(MINI) |
| 37 | + const y1 = Array(budget + 1).fill(MINI) |
| 38 | + for (let i = 0; i <= budget; i++) { |
| 39 | + if (x[i] > MINI) { |
| 40 | + for (let j = 0; j <= budget - i; j++) { |
| 41 | + if (c0[j] > MINI) { |
| 42 | + x1[i + j] = Math.max(x1[i + j], x[i] + c0[j]) |
| 43 | + } |
| 44 | + } |
| 45 | + } |
| 46 | + } |
| 47 | + for (let i = 0; i <= budget; i++) { |
| 48 | + if (y[i] > MINI) { |
| 49 | + for (let j = 0; j <= budget - i; j++) { |
| 50 | + if (c1[j] > MINI) { |
| 51 | + y1[i + j] = Math.max(y1[i + j], y[i] + c1[j]) |
| 52 | + } |
| 53 | + } |
| 54 | + } |
| 55 | + } |
| 56 | + x = x1 |
| 57 | + y = y1 |
| 58 | + } |
| 59 | + const dp1 = v === 0 ? a : b |
| 60 | + for (let i = 0; i <= budget; i++) { |
| 61 | + dp1[i] = Math.max(x[i], y[i]) |
| 62 | + } |
| 63 | + } |
| 64 | + return [a, b] |
| 65 | + } |
| 66 | +} |
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