From 56306fc00b5fc111d66390f3d8c5bb0d25ffa166 Mon Sep 17 00:00:00 2001
From: wangq635 <643797037@qq.com>
Date: Sat, 7 Sep 2024 19:50:44 +0800
Subject: [PATCH] =?UTF-8?q?=E4=BF=AE=E6=AD=A3=20=E5=8A=A8=E6=80=81?=
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---
 problems/背包理论基础01背包-2.md | 17 ++++++-----------
 1 file changed, 6 insertions(+), 11 deletions(-)

diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md
index cd8f317c..830958ef 100644
--- a/problems/背包理论基础01背包-2.md
+++ b/problems/背包理论基础01背包-2.md
@@ -294,23 +294,18 @@ public class Main {
 
 ```python
 n, bagweight = map(int, input().split())
-
 weight = list(map(int, input().split()))
 value = list(map(int, input().split()))
 
-dp = [[0] * (bagweight + 1) for _ in range(n)]
+dp = [0] * (bagweight + 1)  # 创建一个动态规划数组dp，初始值为0
 
-for j in range(weight[0], bagweight + 1):
-    dp[0][j] = value[0]
+dp[0] = 0  # 初始化dp[0] = 0,背包容量为0，价值最大为0
 
-for i in range(1, n):
-    for j in range(bagweight + 1):
-        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])
+for i in range(n):  # 应该先遍历物品，如果遍历背包容量放在上一层，那么每个dp[j]就只会放入一个物品
+    for j in range(bagweight, weight[i]-1, -1):  # 倒序遍历背包容量是为了保证物品i只被放入一次
+        dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
 
-print(dp[n - 1][bagweight])
+print(dp[bagweight])
 
 ```
 ### Go