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krahets
2024-03-21 04:22:07 +08:00
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8
docs/chapter_dynamic_programming/knapsack_problem.md
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@ -802,8 +802,8 @@ $$
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: cap + 1), count: n + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for c in stride(from: 1, through: cap, by: 1) {
for i in 1 ... n {
for c in 1 ... cap {
if wgt[i - 1] > c {
// 若超过背包容量，则不选物品 i
dp[i][c] = dp[i - 1][c]
@ -1191,9 +1191,9 @@ $$
// 初始化 dp 表
var dp = Array(repeating: 0, count: cap + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for i in 1 ... n {
// 倒序遍历
for c in stride(from: cap, through: 1, by: -1) {
for c in (1 ... cap).reversed() {
if wgt[i - 1] <= c {
// 不选和选物品 i 这两种方案的较大值
dp[c] = max(dp[c], dp[c - wgt[i - 1]] + val[i - 1])