Merge pull request #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

problems/背包理论基础01背包-1.md

@@ -437,6 +437,58 @@ print(dp[n - 1][bagweight])
 ### Go
 
 ```go
+package main
+
+import (
+    "fmt"
+)
+
+func main() {
+    var n, bagweight int // bagweight代表行李箱空间
+    fmt.Scan(&n, &bagweight)
+
+    weight := make([]int, n) // 存储每件物品所占空间
+    value := make([]int, n)  // 存储每件物品价值
+
+    for i := 0; i < n; i++ {
+        fmt.Scan(&weight[i])
+    }
+    for j := 0; j < n; j++ {
+        fmt.Scan(&value[j])
+    }
+    // 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
+    dp := make([][]int, n)
+    for i := range dp {
+        dp[i] = make([]int, bagweight + 1)
+    }
+
+    // 初始化, 因为需要用到dp[i - 1]的值
+    // j < weight[0]已在上方被初始化为0
+    // j >= weight[0]的值就初始化为value[0]
+    for j := weight[0]; j <= bagweight; j++ {
+        dp[0][j] = value[0]
+    }
+
+    for i := 1; i < n; i++ { // 遍历科研物品
+        for j := 0; j <= bagweight; j++ { // 遍历行李箱容量
+            if j < weight[i] {
+                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
+            } else {
+                dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
+            }
+        }
+    }
+
+    fmt.Println(dp[n-1][bagweight])
+}
+
+func max(x, y int) int {
+    if x > y {
+        return x
+    }
+    return y
+}
+
 ```
 
 ### Javascript

problems/背包理论基础01背包-2.md

@@ -322,46 +322,41 @@ import (
 )
 
 func main() {
-    var n, bagweight int
+    // 读取 M 和 N
-    fmt.Scan(&n, &bagweight)
+    var M, N int
+    fmt.Scan(&M, &N)
 
-    weight := make([]int, n)
+    costs := make([]int, M)
-    value := make([]int, n)
+    values := make([]int, M)
 
-    for i := 0; i < n; i++ {
+    for i := 0; i < M; i++ {
-        fmt.Scan(&weight[i])
+        fmt.Scan(&costs[i])
     }
-    for j := 0; j < n; j++ {
+    for j := 0; j < M; j++ {
-        fmt.Scan(&value[j])
+        fmt.Scan(&values[j])
     }
 
-    dp := make([][]int, n)
+    // 创建一个动态规划数组dp，初始值为0
-    for i := range dp {
+    dp := make([]int, N + 1)
-        dp[i] = make([]int, bagweight+1)
-    }
 
-    for j := weight[0]; j <= bagweight; j++ {
+    // 外层循环遍历每个类型的研究材料
-        dp[0][j] = value[0]
+    for i := 0; i < M; i++ {
-    }
+        // 内层循环从 N 空间逐渐减少到当前研究材料所占空间
-
+        for j := N; j >= costs[i]; j-- {
-    for i := 1; i < n; i++ {
+            // 考虑当前研究材料选择和不选择的情况，选择最大值
-        for j := 0; j <= bagweight; j++ {
+            dp[j] = max(dp[j], dp[j-costs[i]] + values[i])
-            if j < weight[i] {
-                dp[i][j] = dp[i-1][j]
-            } else {
-                dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
-            }
         }
     }
 
-    fmt.Println(dp[n-1][bagweight])
+    // 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
+    fmt.Println(dp[N])
 }
 
-func max(a, b int) int {
+func max(x, y int) int {
-    if a > b {
+    if x > y {
-        return a
+        return x
     }
-    return b
+    return y
 }
 
 ```