## 背包问题-最小价值-空间恰好$j$

### 一、$01$背包
求价值最小值：初始化$f[0][0] = 0$, 其余是$INF$

例子：给你一堆物品，每个物品有一定的体积和对应的价值，每个物品只能选一个，求 <font color='blue' size=4><b>总体积恰好是$m$</b></font>的 <font color='red' size=4><b>最小价值</b></font>

输入
```c++
4 5
1 2
2 4
3 4
4 5
```
输出
```c++
7
```

#### 1、二维
```c++
#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;

int n, m;
int f[N][N];

int main() {
    //文件输入
    freopen("QiaHao_01_Min.in", "r", stdin);
    scanf("%d %d", &n, &m);
    memset(f, 0x3f, sizeof f); //预求最小，先设最大
    f[0][0] = 0;              //前0个物品中选择，装满0个空间，获得的最小值为0

    for (int i = 1; i <= n; i++) {
        int v, w;
        scanf("%d %d", &v, &w);
        for (int j = 0; j <= m; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= v) f[i][j] = min(f[i][j], f[i - 1][j - v] + w);
        }
    }
    printf("%d\n", f[n][m]);
    return 0;
}
```

#### 2、一维
```c++
#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;

int n, m;
int f[N];

int main() {
    //文件输入
    freopen("QiaHao_01_Min.in", "r", stdin);
    scanf("%d %d", &n, &m);
    memset(f, 0x3f, sizeof f);
    f[0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        scanf("%d %d", &v, &w);
        for (int j = m; j >= v; j--)
            f[j] = min(f[j], f[j - v] + w);
    }
    printf("%d\n", f[m]);
    return 0;
}
```

### 二、完全背包

求价值最小值：初始化$f[0][0] = 0$, 其余是$INF$

例子：给你一堆物品，每个物品有一定的体积和对应的价值，每个物品可以选无数多个，求 <font color='blue' size=4><b>总体积恰好是$m$</b></font>的 <font color='red' size=4><b>最小价值</b></font>

输入
```c++
4 5
1 2
2 4
3 4
4 5
```
输出
```c++
7
```
#### 1、二维
```c++
#include <bits/stdc++.h>

using namespace std;
const int N = 110;
const int INF = 0x3f3f3f3f;

int n, m;
int f[N][N];

int main() {
    scanf("%d %d", &n, &m);

    memset(f, 0x3f, sizeof f);
    f[0][0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        scanf("%d %d", &v, &w);
        for (int j = 0; j <= m; j++) {
            f[i][j] = f[i - 1][j];
            if (j >= v) f[i][j] = min(f[i][j], f[i][j - v] + w);
        }
    }
    printf("%d\n", f[n][m]);
    return 0;
}
```
#### 2、一维
```c++
#include <bits/stdc++.h>

using namespace std;
const int N = 110, INF = 0x3f3f3f3f;

int n, m;
int f[N];

int main() {
    //文件输入
    freopen("QiaHao_WQ_Min.in", "r", stdin);
    scanf("%d %d", &n, &m);

    memset(f, 0x3f, sizeof f);
    f[0] = 0;

    for (int i = 1; i <= n; i++) {
        int v, w;
        scanf("%d %d", &v, &w);
        for (int j = v; j <= m; j++)
            f[j] = min(f[j], f[j - v] + w);
    }
    printf("%d\n", f[m]);
    return 0;
}
```