Multiple backpacks!

Yes N Items And a Capacity of V The backpack . The first i There are at most Mi Pieces available , The space consumed by each piece is Ci , The value is Wi . Find out which items can be loaded into the backpack to make the space consumed by these items The total does not exceed the capacity of the backpack , And The sum of value is the greatest .


It makes no difference , This turns into a 01 It's a knapsack problem , And only once per item .

Code

public void testMultiPack1(){
    
    //  Version of a ： Change the number of items to 01 Knapsack format 
    List<Integer> weight = new ArrayList<>(Arrays.asList(1, 3, 4));
    List<Integer> value = new ArrayList<>(Arrays.asList(15, 20, 30));
    List<Integer> nums = new ArrayList<>(Arrays.asList(2, 3, 2));
    int bagWeight = 10;

    for (int i = 0; i < nums.size(); i++) {
    
        while (nums.get(i) > 1) {
     //  Expand the item into i
            weight.add(weight.get(i));
            value.add(value.get(i));
            nums.set(i, nums.get(i) - 1);
        }
    }

    int[] dp = new int[bagWeight + 1];
    for(int i = 0; i < weight.size(); i++) {
     //  Traverse the items 
        for(int j = bagWeight; j >= weight.get(i); j--) {
     //  Traverse the backpack capacity 
            dp[j] = Math.max(dp[j], dp[j - weight.get(i)] + value.get(i));
        }
        System.out.println(Arrays.toString(dp));
    }
}