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krahets
2024-03-21 04:22:07 +08:00
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docs/chapter_dynamic_programming/unbounded_knapsack_problem.md
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@ -166,8 +166,8 @@ $$
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: cap + 1), count: n + 1)
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for c in stride(from: 1, through: cap, by: 1) {
for i in 1 ... n {
for c in 1 ... cap {
if wgt[i - 1] > c {
// 若超过背包容量，则不选物品 i
dp[i][c] = dp[i - 1][c]
@ -506,8 +506,8 @@ $$
// 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 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[c] = dp[c]
@ -882,12 +882,12 @@ $$
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: amt + 1), count: n + 1)
// 状态转移：首行首列
for a in stride(from: 1, through: amt, by: 1) {
for a in 1 ... amt {
dp[0][a] = MAX
}
// 状态转移：其余行和列
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
for i in 1 ... n {
for a in 1 ... amt {
if coins[i - 1] > a {
// 若超过目标金额，则不选硬币 i
dp[i][a] = dp[i - 1][a]
@ -1300,8 +1300,8 @@ $$
var dp = Array(repeating: MAX, count: amt + 1)
dp[0] = 0
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
for i in 1 ... n {
for a in 1 ... amt {
if coins[i - 1] > a {
// 若超过目标金额，则不选硬币 i
dp[a] = dp[a]
@ -1660,12 +1660,12 @@ $$
// 初始化 dp 表
var dp = Array(repeating: Array(repeating: 0, count: amt + 1), count: n + 1)
// 初始化首列
for i in stride(from: 0, through: n, by: 1) {
for i in 0 ... n {
dp[i][0] = 1
}
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
for i in 1 ... n {
for a in 1 ... amt {
if coins[i - 1] > a {
// 若超过目标金额，则不选硬币 i
dp[i][a] = dp[i - 1][a]
@ -2002,8 +2002,8 @@ $$
var dp = Array(repeating: 0, count: amt + 1)
dp[0] = 1
// 状态转移
for i in stride(from: 1, through: n, by: 1) {
for a in stride(from: 1, through: amt, by: 1) {
for i in 1 ... n {
for a in 1 ... amt {
if coins[i - 1] > a {
// 若超过目标金额，则不选硬币 i
dp[a] = dp[a]