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Commit 0114d33

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Merge pull request youngyangyang04#2681 from markwang1992/01bag
0-1背包理论基础(二)增加Go一维dp数组,二维dp数组移至0-1背包理论基础(一),增加相应注释
2 parents 0f74e69 + bb2803c commit 0114d33

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2 files changed

+76
-29
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2 files changed

+76
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‎problems/背包理论基础01背包-1.md‎

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Original file line numberDiff line numberDiff line change
@@ -437,6 +437,58 @@ print(dp[n - 1][bagweight])
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### Go
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```go
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package main
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import (
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"fmt"
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)
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func main() {
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var n, bagweight int // bagweight代表行李箱空间
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fmt.Scan(&n, &bagweight)
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weight := make([]int, n) // 存储每件物品所占空间
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value := make([]int, n) // 存储每件物品价值
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for i := 0; i < n; i++ {
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fmt.Scan(&weight[i])
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}
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for j := 0; j < n; j++ {
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fmt.Scan(&value[j])
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}
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// dp数组, dp[i][j]代表行李箱空间为j的情况下,从下标为[0, i]的物品里面任意取,能达到的最大价值
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dp := make([][]int, n)
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for i := range dp {
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dp[i] = make([]int, bagweight + 1)
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}
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// 初始化, 因为需要用到dp[i - 1]的值
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// j < weight[0]已在上方被初始化为0
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// j >= weight[0]的值就初始化为value[0]
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for j := weight[0]; j <= bagweight; j++ {
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dp[0][j] = value[0]
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}
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for i := 1; i < n; i++ { // 遍历科研物品
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for j := 0; j <= bagweight; j++ { // 遍历行李箱容量
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if j < weight[i] {
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dp[i][j] = dp[i-1][j] // 如果装不下这个物品,那么就继承dp[i - 1][j]的值
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} else {
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dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
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}
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}
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}
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fmt.Println(dp[n-1][bagweight])
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}
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func max(x, y int) int {
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if x > y {
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return x
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}
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return y
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}
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```
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### Javascript

‎problems/背包理论基础01背包-2.md‎

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Original file line numberDiff line numberDiff line change
@@ -313,7 +313,7 @@ for i in range(1, n):
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print(dp[n - 1][bagweight])
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```
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### Go
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### Go
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```go
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package main
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@@ -322,46 +322,41 @@ import (
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)
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func main() {
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var n, bagweight int
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fmt.Scan(&n, &bagweight)
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// 读取 M 和 N
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var M, N int
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fmt.Scan(&M, &N)
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weight := make([]int, n)
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value := make([]int, n)
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costs := make([]int, M)
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values := make([]int, M)
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331-
for i := 0; i < n; i++ {
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fmt.Scan(&weight[i])
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}
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for j := 0; j < n; j++ {
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fmt.Scan(&value[j])
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for i := 0; i < M; i++ {
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fmt.Scan(&costs[i])
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}
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dp := make([][]int, n)
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for i := range dp {
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dp[i] = make([]int, bagweight+1)
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for j := 0; j < M; j++ {
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fmt.Scan(&values[j])
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}
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343-
for j := weight[0]; j <= bagweight; j++ {
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dp[0][j] = value[0]
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}
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// 创建一个动态规划数组dp,初始值为0
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dp := make([]int, N + 1)
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for i := 1; i < n; i++ {
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for j := 0; j <= bagweight; j++ {
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if j < weight[i] {
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dp[i][j] = dp[i-1][j]
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} else {
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dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
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}
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// 外层循环遍历每个类型的研究材料
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for i := 0; i < M; i++ {
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// 内层循环从 N 空间逐渐减少到当前研究材料所占空间
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for j := N; j >= costs[i]; j-- {
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// 考虑当前研究材料选择和不选择的情况,选择最大值
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dp[j] = max(dp[j], dp[j-costs[i]] + values[i])
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}
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}
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357-
fmt.Println(dp[n-1][bagweight])
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// 输出dp[N],即在给定 N 行李空间可以携带的研究材料最大价值
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fmt.Println(dp[N])
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}
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360-
func max(a, b int) int {
361-
if a > b {
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return a
355+
func max(x, y int) int {
356+
if x > y {
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return x
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}
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return b
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return y
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}
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```

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