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python/TangDou/AcWing/BeiBao/【总结】最大价值-空间至多j.md

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## 背包问题-最大价值-空间至多$j$
<font color='red' size=5><b>最简背包问题,恰好和至少都更复杂,这个是最原始的版本。</b></font>
### 一、01背包
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>总体积不超过$m$</b></font><font color='red' size=4><b>最大价值</b></font>
输入
```c++
4 5
1 2
2 4
3 4
4 5
```
输出
```c++
8
```
#### 1、二维
```c++
#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int n, m;
int f[N][N];
int main() {
scanf("%d %d", &n, &m);
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] = max(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;
int n, m;
int f[N];
int main() {
scanf("%d %d", &n, &m);
for (int i = 1; i <= n; i++) {
int v, w;
scanf("%d %d", &v, &w);
for (int j = m; j >= v; j--)
f[j] = max(f[j], f[j - v] + w);
}
printf("%d\n", f[m]);
return 0;
}
```
### 二、完全背包
```c++
4 5
1 2
2 4
3 4
4 5
```
答案
```c++
10
```
#### 1、二维
```c++
#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, m;
int f[N][N];
int main() {
scanf("%d %d", &n, &m);
for (int i = 1; i <= n; i++) {
int v, w;
scanf("%d %d", &v, &w);
for (int j = 1; j <= m; j++) {
f[i][j] = f[i - 1][j];
if (j >= v) f[i][j] = max(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 = 1010;
int n, m;
int f[N];
// 完全背包问题
int main() {
scanf("%d %d", &n, &m);
for (int i = 1; i <= n; i++) {
int v, w;
scanf("%d %d", &v, &w);
for (int j = v; j <= m; j++) //一个一个加上来,求一个最大值
f[j] = max(f[j], f[j - v] + w);
}
printf("%d\n", f[m]);
return 0;
}
```