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## 背包问题
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### 一、$01$背包基础题
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**[$AcWing$ $2$. $01$背包问题](https://www.acwing.com/problem/content/2/)**
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**[$AcWing$ $423$. 采药](https://www.acwing.com/problem/content/425/)**
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**[$AcWing$ $1024$. 装箱问题](https://www.acwing.com/problem/content/1026/)**
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二维状态表示
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110;
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const int M = 1010;
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int n, m;
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int w[N], v[N];
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int f[N][M];
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int main() {
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cin >> m >> n;
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for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
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for (int i = 1; i <= n; i++)
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for (int j = 1; j <= m; j++) {
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f[i][j] = f[i - 1][j]; // 不选
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if (j >= v[i])
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f[i][j] = max(f[i][j], f[i - 1][j - v[i]] + w[i]); // 选
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}
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printf("%d\n", f[n][m]);
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return 0;
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}
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```
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一维状态表示
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 1010;
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int n, m;
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int v[N], w[N];
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int f[N];
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int main() {
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cin >> m >> n;
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for (int i = 1; i <= n; i++) cin >> v[i] >> w[i];
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// 01背包模板
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for (int i = 1; i <= n; i++)
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for (int j = m; j >= v[i]; j--)
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f[j] = max(f[j], f[j - v[i]] + w[i]);
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printf("%d\n", f[m]);
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return 0;
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}
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```
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### 二、二维费用$01$背包问题
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**[$AcWing$ $1022$. 宠物小精灵之收服](https://www.acwing.com/problem/content/1024/)**
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**[$AcWing$ $8$. 二维费用的背包问题](https://www.cnblogs.com/littlehb/p/15684961.html)**
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```cpp {.line-numbers}
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#include <bits/stdc++.h>
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using namespace std;
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const int N = 110; // 野生小精灵的数量
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const int M1 = 1010; // 小智的精灵球数量
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const int M2 = 510; // 皮卡丘的体力值
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int n, m1, m2;
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int f[M1][M2]; // 一维：精灵球数量，二维：皮卡丘的体力值，值：抓到的小精灵数量最大值
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int main() {
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cin >> m1 >> m2 >> n;
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m2--; // 留一滴血
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// 二维费用01背包
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// 降维需要将体积1、体积2倒序枚举
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for (int i = 1; i <= n; i++) {
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int v1, v2;
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cin >> v1 >> v2;
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for (int j = m1; j >= v1; j--)
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for (int k = m2; k >= v2; k--)
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f[j][k] = max(f[j][k], f[j - v1][k - v2] + 1); // 获利就是多了一个小精灵
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}
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// 最多收服多少个小精灵[在消耗精灵球、血极限的情况下，肯定抓的是最多的，这不废话吗]
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printf("%d ", f[m1][m2]);
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// 找到满足最大价值的所有状态里，第二维费用消耗最少的
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int cost = m2;
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for (int i = 0; i <= m2; i++) // 如果一个都不收服，则体力消耗最少，消耗值为0
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if (f[m1][i] == f[m1][m2])
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cost = min(cost, i);
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// 收服最多个小精灵时皮卡丘的剩余体力值最大是多少
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printf("%d\n", m2 + 1 - cost);
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return 0;
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}
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```
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**总结**
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- $01$背包，还是背一维的形式比较好，一来代码更短，二来空间更省，倒序就完了。
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- 二维费用的$01$背包，简化版本的$01$背包模板就有了用武之地，因为三维数组可能会爆内存。 |
