From c6aeaa512a66f635cf73cfe14496e25fcc574ea7 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E9=BB=84=E6=B5=B7?= <10402852@qq.com>
Date: Tue, 12 Mar 2024 16:47:09 +0800
Subject: [PATCH] 'commit'

---
 TangDou/AcWing/BeiBao/1020.md    | 30 ++++++------------------------
 TangDou/AcWing/BeiBao/1020_1.cpp |  6 +++---
 2 files changed, 9 insertions(+), 27 deletions(-)

diff --git a/TangDou/AcWing/BeiBao/1020.md b/TangDou/AcWing/BeiBao/1020.md
index 501894d..659f00c 100644
--- a/TangDou/AcWing/BeiBao/1020.md
+++ b/TangDou/AcWing/BeiBao/1020.md
@@ -87,48 +87,30 @@ $f(i,j,k)$ 表示从前$i$个物品中选，且花费$1$**不少于**$j$,花费$
 
 **答**：不需要，都可以满足要求了，即$j=0,k=0$,也就是$f[i-1][0][0]+w$,而对于一个无欲无求的$f[i-1][0][0]$自然是等于$0$,也就是$f[i][j][k]=w$
 
-### 三、三维朴素解法
+### 三、三维解法
 ```cpp {.line-numbers}
 #include <bits/stdc++.h>
-
 using namespace std;
 const int N = 1010;
 const int M = 110;
+int n, m, m1, m2;
 int f[N][M][M];
-int n, m1, m2;
 
-//二维费用01背包-不少于维度费用，求最小代价
 int main() {
-    //注意次序
-    scanf("%d %d %d", &m1, &m2, &n);
-    //求最小值
+    cin >> m1 >> m2 >> n;
     memset(f, 0x3f, sizeof f);
     f[0][0][0] = 0;
 
     for (int i = 1; i <= n; i++) {
         int v1, v2, w;
-        scanf("%d %d %d", &v1, &v2, &w);
+        cin >> v1 >> v2 >> w;
         for (int j = 0; j <= m1; j++)
             for (int k = 0; k <= m2; k++) {
-                //不选择i号物品
                 f[i][j][k] = f[i - 1][j][k];
-                //分情况讨论
-
-                //物品i加上就够一维使用，此时，只关心二维情况即可
-                if (j - v1 < 0 && k - v2 >= 0)
-                    f[i][j][k] = min(f[i][j][k], f[i - 1][0][k - v2] + w);
-                //物品i加上就够二维使用，此时，只关心一维情况即可
-                else if (j - v1 >= 0 && k - v2 < 0)
-                    f[i][j][k] = min(f[i][j][k], f[i - 1][j - v1][0] + w);
-                //如果选择了i号物品，两个维度直接拉满，那么只选择一个i就足够用，它参选的价值是w
-                else if (j - v1 < 0 && k - v2 < 0)
-                    f[i][j][k] = min(f[i][j][k], w);
-                else
-                    //正常递推
-                    f[i][j][k] = min(f[i][j][k], f[i - 1][j - v1][k - v2] + w);
+                f[i][j][k] = min(f[i - 1][j][k], f[i - 1][max(0, j - v1)][max(0, k - v2)] + w);
             }
     }
-    printf("%d\n", f[n][m1][m2]);
+    cout << f[n][m1][m2] << endl;
     return 0;
 }
 ```
diff --git a/TangDou/AcWing/BeiBao/1020_1.cpp b/TangDou/AcWing/BeiBao/1020_1.cpp
index 10f958c..f207c48 100644
--- a/TangDou/AcWing/BeiBao/1020_1.cpp
+++ b/TangDou/AcWing/BeiBao/1020_1.cpp
@@ -11,12 +11,12 @@ int main() {
     f[0][0][0] = 0;
 
     for (int i = 1; i <= n; i++) {
-        int u, v, w;
-        cin >> u >> v >> w;
+        int v1, v2, w;
+        cin >> v1 >> v2 >> w;
         for (int j = 0; j <= m1; j++)
             for (int k = 0; k <= m2; k++) {
                 f[i][j][k] = f[i - 1][j][k];
-                f[i][j][k] = min(f[i - 1][j][k], f[i - 1][max(0, j - u)][max(0, k - v)] + w);
+                f[i][j][k] = min(f[i - 1][j][k], f[i - 1][max(0, j - v1)][max(0, k - v2)] + w);
             }
     }
     cout << f[n][m1][m2] << endl;