> For the complete documentation index, see [llms.txt](https://junnie.gitbook.io/nine-chapter/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://junnie.gitbook.io/nine-chapter/9.dynamic-programming/92backpack.md).

# Backpack I\~VI

## 1.(92)Backpack I(单次选择+最大体积)

Given\_n\_items with size Ai, an integer\_m\_denotes the size of a backpack. How full you can fill this backpack?

### Notice

You can not divide any item into small pieces.

**Example**

If we have`4`items with size`[2, 3, 5, 7]`, the backpack size is 11, we can select`[2, 3, 5]`, so that the max size we can fill this backpack is`10`. If the backpack size is`12`. we can select`[2, 3, 7]`so that we can fulfill the backpack.

You function should return the max size we can fill in the given backpack.

[**Challenge**](http://lintcode.com/en/problem/backpack/#challenge)

O(n x m) time and O(m) memory.

O(n x m) memory is also acceptable if you do not know how to optimize memory.

## Code

动规经典题目，用数组dp\[i]表示书包空间为i的时候能装的A物品最大容量。两次循环，外部遍历数组A，内部反向遍历数组dp，若j即背包容量大于等于物品体积A\[i]，则取前i-1次循环求得的最大容量dp\[j]，和背包体积为j-A\[i]时的最大容量dp\[j-A\[i]]与第i个物品体积A\[i]之和即dp\[j-A\[i]]+A\[i]的较大值，作为本次循环后的最大容量dp\[i]。

注意dp\[]的空间要给m+1，因为我们要求的是第m+1个值dp\[m]，否则会抛出OutOfBoundException。

```
public int backPack(int m, int[] A) {

        int[] fill=new int[m+1];
        fill[0]=0;
        for(int i=0;i<A.length;i++){
            for(int j=m;j>0;j--){
                if(j>=A[i]){
                    fill[j]=Math.max(fill[j],fill[j-A[i]]+A[i]);
                }
            }
        }
        return fill[m];
    }
```

## 2.(125)Backpack II (单次选择+最大价值)

Given*n\_items with size Aiand value Vi, and a backpack with size\_m*. What's the maximum value can you put into the backpack?

### Notice

You cannot divide item into small pieces and the total size of items you choose should smaller or equal to m.

**Example**

Given 4 items with size`[2, 3, 5, 7]`and value`[1, 5, 2, 4]`, and a backpack with size`10`. The maximum value is`9`.

[**Challenge**](http://lintcode.com/en/problem/backpack-ii/#challenge)

O(n x m) memory is acceptable, can you do it in O(m) memory?

## Code

和BackPack I基本一致。依然是以背包空间为限制条件，所不同的是dp\[j]取的是价值较大值，而非体积较大值。所以只要把dp\[j-A\[i]]+A\[i]换成dp\[j-A\[i]]+V\[i]就可以了。

```
 public int backPackII(int m, int[] A, int V[]) {
        int[] value=new int[m+1];
        value[0]=0;
        for(int i=0;i<A.length;i++){
            for(int j=m;j>0;j--){
                if(j>=A[i]){
                    value[j]=Math.max(value[j],value[j-A[i]]+V[i]);
                }
            }
        }
        return value[m];    
    }
```


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