01背包问题暴力解法（回溯法）和经典解法

暴力解法（回溯法）

import java.util.Arrays;
import java.util.Scanner;

public class Main {
	private final static int  N = 999;
	public static int SumValue = 0;
	public static int SumWeight = 0;
	public static int OptimalValue = 0;
	public static int OptimalSolution[] = new int[N];
	public static int w[] = new int[N];
	public static int v[] = new int[N];
	public static int x[] = new int[N];
	public static int n;
	public static int c;
	
	public static void  backTracking(int t) {
		if(t>n) {
			if(SumValue > OptimalValue) {
				OptimalValue = SumValue;
				for(int i=1;i<=n;i++)
					OptimalSolution[i]=x[i];
			}
		}else {
			for(int i=0;i<=1;i++) {
				x[t]=i;
				if(i==0) {
					backTracking(t+1);
				}else {
					if(SumValue+w[t]<=c) {
						SumWeight += w[t];
						SumValue += v[t];
						backTracking(t+1);
						SumValue -= v[t];
						SumWeight -= w[t];
					}
				}
			}
		}
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc =new Scanner(System.in);
		n = sc.nextInt();
		Arrays.fill(w, 0);
		Arrays.fill(v, 0);
		Arrays.fill(x, 0);
		c = sc.nextInt();
		for(int i=1;i<=n;i++)
		{
			int temp = sc.nextInt();
			w[i]=temp;
			temp = sc.nextInt();
			v[i]=temp;
		}
		backTracking(1);
		System.out.print(OptimalValue);
		
	}

}

二维数组（经典解法） 这里dp[i][j]表示含义是取[0,i]个物品且容量不超过j的最大价值为dp[i][j]

import java.nio.file.Watchable;
import java.util.Scanner;

public class 经典解法Test {
	public static void main(String[] args) {
		int n;
		int N = 999;
		int[][] dp =new int[N][N];
		Scanner scanner=new Scanner(System.in);
		n=scanner.nextInt();
		int c;
		c=scanner.nextInt();
		int[] w=new int[N];
		int[] v = new int[N];
		for(int i=0;i<n;i++) {
			int temp=scanner.nextInt();
			w[i]=temp;
			temp=scanner.nextInt();
			v[i]=temp;
		}
		for(int i=0;i<n;i++)
			dp[i][0]=0;
		for(int i=w[0];i<=c;i++)
			dp[0][i]=v[0];
		for(int i=1;i<n;i++) {
			for(int j=c;j>=0;j--) {
				if(j>=w[i]) {
					dp[i][j]=Math.max(dp[i-1][j], dp[i-1][j-w[i]]+v[i]);
				}else {
					dp[i][j]=dp[i-1][j];
				}
			}
		}
		System.out.println(dp[n-1][c]);
	}
}

一维数组（经典解法）dp[j] 含义是容量不超过j且最大价值为dp[j]

import java.util.Scanner;

public class 一维数组 {
	public static void main(String[] args) {
		int N=999;
		int n,c;
		int[] w= new int[N];
		int[] v=new int[N];
		int[] dp=new int[N];
		Scanner scanner=new Scanner(System.in);
		n=scanner.nextInt();
		c=scanner.nextInt();
		for(int i=0;i<n;i++) {
			int temp=scanner.nextInt();
			w[i]=temp;
			temp=scanner.nextInt();
			v[i]=temp;
		}
		dp[0]=0;
		for(int i=0;i<n;i++) {
			for(int j=c;j>=w[i];j--)
				dp[j]=Math.max(dp[j], dp[j-w[i]]+v[i]);
		}
		System.out.println(dp[c]);
		
	}
}