01背包 Java

① 记忆化搜索解法：

import java.util.*;
import java.io.*;

public class Main {
	static int n, m;
	static int[] v, w;
	static int[][] memory; // 记忆化数组

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		n = Integer.parseInt(st.nextToken());
		m = Integer.parseInt(st.nextToken());
		v = new int[n + 1];
		w = new int[n + 1];
		memory = new int[n + 1][m + 1];
		for (int i = 1; i <= n; i++) {
			st = new StringTokenizer(br.readLine());
			v[i] = Integer.parseInt(st.nextToken());
			w[i] = Integer.parseInt(st.nextToken());
		}
		System.out.println(dfs(1, 0));
	}

	static int dfs(int i, int vSum) {
		if (i > n) return 0;
		if (memory[i][vSum] != 0) return memory[i][vSum]; // 直接返回存储的值，跳过重复计算
		int reject = dfs(i + 1, vSum); // 左分支，不选当前物品
		int accept = 0; // 右分支，选中当前物品
		if (vSum + v[i] <= m)
			accept = dfs(i + 1, vSum + v[i]) + w[i]; // 要加上物品价值
		return memory[i][vSum] = Math.max(reject, accept); // 存储当前的值
	}
}

② dp解法（更优化）：

不能选(背包空间不足)：继承前面，即dp[i - 1][j]。

能选(背包容量足够)：获得该层价值w[i]，同时失去对应背包容量v[i]，然后获得选与不选其中价值更大的一个。

import java.util.*;
import java.io.*;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		int[] v = new int[1005];
		int[] w = new int[1005];
		int[][] dp = new int[1005][1005];
		for (int i = 1; i <= n; i++) {
			st = new StringTokenizer(br.readLine());
			v[i] = Integer.parseInt(st.nextToken());
			w[i] = Integer.parseInt(st.nextToken());
		}
		for (int i = 1; i <= n; i++) {
			for (int j = 0; j <= m; j++) {
				if (j < v[i]) dp[i][j] = dp[i - 1][j];
				else dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - v[i]] + w[i]);
			}
		}
		System.out.println(dp[n][m]);
	}
}

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