Multiple knapsack problem

Concept ： Yes N Class 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 values is the largest .

Multiple backpacks and 01 Backpacks are very much like , Why and 01 Backpacks are like ？

Each item has at most Mi Pieces available , hold Mi The pieces are spread out , It's really just a 01 It's a knapsack problem .

 for example ：

 The maximum weight of the backpack is 10.

 Item is ：

        weight   value   Number 
 goods 0	1	15	 2
 goods 1	3	20	 3
 goods 2	4	30	 2
 What's the maximum value of the items a backpack can carry ？

Is it different from ？

        weight   value   Number 
 goods 0	1	15	 1
 goods 0	1	15	 1
 goods 1	3	20	 1
 goods 1	3	20	 1
 goods 1	3	20	 1
 goods 2	4	30	 1
 goods 2	4	30	 1

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

The code to implement multiple knapsack in this way is as follows ：

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));
    }
}

There's another way to do it , It is to put the number of traversals of each commodity in 01 I'm going through it in my backpack .

class Solution {
     public boolean wordBreak(String s, List<String> wordDict) {
         List<Integer> weight = new ArrayList<>(Arrays.aslist(1, 3, 4));
         List<Integer> price = new ArrayList<>(Arrays.aslist(15, 20, 20));
         List<Integer> nums = new ArrayList<>(Arrays.aslist(2, 3, 2));

         int bagWeight = 10;
        //  for(int i = 0; i < nums.size(); i++){
        //      if(nums[i] > 0){
        //          weight.add(weight.get(i));
        //          price.add(price.get(i));
        //          nums[i]--;
        //      }
        //  }
         int[] dp = new int[bagWeight+1];
         for(int i = 0; i < weight.size();i++){
             for(int j = bagWeight; j >= weight.get(i); j--){
                 for(int k = 1; k <= nums[i] && j- k * weight[i] >= 0; k++){
                     dp[j] = Math.max(dp[j], dp[j-k * weight[i]] + k * price[i]);
                 }               
             }
         }
         return dp[bagWeight];
     }
}