題目網站：https://www.acwing.com/problem/content/description/2/
dp[i][j]表示1前i件物品放入體積為j的揹包裡可以獲得的最大價值。

1. 0-1揹包（AcWing2）

import java.util.Scanner;

public class Main{
    public static void main(String[] args) throws Exception {
        Scanner reader = new Scanner(System.in);
        int N = reader.nextInt();
        int V = reader.nextInt();
        int[] v = new int[N + 1] ;
        int[] w = new int[N + 1] ;

        for (int i=1 ; i <= N ; i++){
            v[i] = reader.nextInt();
            w[i] = reader.nextInt();
        }
        reader.close() ;
        int[][] dp = new int[N+1][V+1];
        dp[0][0] = 0;
        for(int i = 1; i <= N; i++){
            for(int j = 0; j <= V; j++){
                if(j >= v[i]){
                    dp[i][j] = Math.max(dp[i-1][j], dp[i-1][j-v[i]] + w[i]);
                }else{
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        System.out.println(dp[N][V]);
    }
}
 

2. 完全揹包（AcWing3）

import java.util.Scanner;

public class Main{
    public static void main(String[] args) throws Exception {
        Scanner reader = new Scanner(System.in);
        int N = reader.nextInt();
        int V = reader.nextInt();
        int[] v = new int[N + 1] ;
        int[] w = new int[N + 1] ;

        for (int i=1 ; i <= N ; i++){
            v[i] = reader.nextInt();
            w[i] = reader.nextInt();
        }
        reader.close() ;
        int[][] dp = new int[N+1][V+1];
        dp[0][0] = 0;
        for(int i = 1; i <= N; i++){
            for(int j = 0; j <= V; j++){
                if(j >= v[i]){
                    //此處是和0-1揹包問題唯一的不同之處,參考了《演算法筆記》上的內容
                    dp[i][j] = Math.max(dp[i-1][j], dp[i][j-v[i]] + w[i]);
                }else{
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        System.out.println(dp[N][V]);
    }
}