题目描述

有 N 种物品和一个容量是 V 的背包。

第 i 种物品最多有 si 件，每件体积是 vi，价值是 wi。

求解将哪些物品装入背包，可使物品体积总和不超过背包容量，且价值总和最大。
输出最大价值。

输入格式
第一行两个整数，N，V (0<N≤1000, 0<V≤20000)，用空格隔开，分别表示物品种数和背包容积。

接下来有 N 行，每行三个整数 vi,wi,si，用空格隔开，分别表示第 i 种物品的体积、价值和数量。

输出格式
输出一个整数，表示最大价值。

数据范围
0<N≤1000
0<V≤20000
0<vi,wi,si≤20000

样例

4 5
1 2 3
2 4 1
3 4 3
4 5 2

10

回顾

(暴力枚举)

对每一种的物品采用分组背包，只能选择1v, 2v,…,kv中的一个

时间复杂度 O(NVS)

Java 代码

import java.util.*;
public class Main {
    public static int backPack3(int N, int V, int[] v, int[] w, int[] s) {
        int[] dp = new int[V + 1];
        for (int i = 0; i < N; i++) {
            // this is wrong, in this case we are acturely doing 0-1 backpack using 0v[i], 1v[i]...s[i]v[i]
            // for (int k = 0; k <= s[i]; k++) {
            //     for (int j = V; j >= k * v[i]; j--) {
            //         dp[j] = Math.max(dp[j], dp[j -  k * v[i]] + k * w[i]);
            //     }
            // }

            // this is correct, only 1 case for k * v[i]
            for (int j = V; j >= 0; j--) {
                for (int k = 0; k <= s[i] && k * v[i] <= j; k++) {
                    dp[j] = Math.max(dp[j], dp[j - k * v[i]] + k * w[i]);
                }
            }
        }
        return dp[V];
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int V = sc.nextInt();
        int[] v = new int[N];
        int[] w = new int[N];
        int[] s = new int[N];
        for (int i = 0; i < N; i++) {
            v[i] = sc.nextInt();
            w[i] = sc.nextInt();
            s[i] = sc.nextInt();
        }
        System.out.println(backPack3(N, V, v, w, s));
    }
}

算法1

(二进制划分)

对每一种的物品数量二进制划分:1v, 2v, 4v, …

时间复杂度 O(NVlog(S))

Java 代码

import java.util.*;
public class Main {
    public static int backPack4(int V, List<Integer> vList, List<Integer> wList) {
        int[] dp = new int[V + 1];
        for (int i = 0; i < vList.size(); i++) {
            for (int j = V; j >= vList.get(i); j--) {
                dp[j] = Math.max(dp[j], dp[j - vList.get(i)] + wList.get(i));
            }    
        }
        return dp[V];
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int V = sc.nextInt();
        List<Integer> vList = new ArrayList<>();
        List<Integer> wList = new ArrayList<>();
        for (int i = 0; i < N; i++) {
            int v = sc.nextInt();
            int w = sc.nextInt();
            int s = sc.nextInt();
            int mask = 1;
            while (s >= mask) {
                vList.add(mask * v);
                wList.add(mask * w);
                s -= mask;
                mask *= 2;
            }
            if (s > 0) {
                vList.add(s * v);
                wList.add(s * w);
            }
        }
        System.out.println(backPack4(V, vList, wList));
    }
}

算法2

(单调队列)

对每一种的物品，设f(i) = dp[j - i * v] + i * w, dp[j] = max(f(1), f(2), …, f(k));
对f序列用单调队列优化

时间复杂度 O(N*V)

Java 代码

import java.util.*;
/*
 *1. for an item with v and w to fill package j, only consider the cases: 
 *   dp[j - v] + w, dp[j - 2 * v] + 2 * w, ... , dp[j - k * v] + k * w (k = min(s[i], j / v));
 *2. f(i) = dp[j - i * v] + i * w, we want to get f(1), f(2), ... ,f(k) dynamicaly, could use Monotone Queue;
 *3. time complexity: O(N * V)
 */
public class Main {
    public static int backPack5(int N, int V, int[] v, int[] w, int[] s) {
        int[] dp = new int[V + 1];
        int[] copy = new int[V + 1];
        int[] queue = new int[V + 1];
        for (int i = 0; i < N; i++) {
            // for index i, copy arr stores the result only using item before i
            for (int c = 0; c <= V; c++) copy[c] = dp[c];
            // status only transfer amone mode v has same result
            for (int j = 0; j < v[i]; j++) {
                // use 2 pointer to simulat monotone queue
                int head = 0;
                int tail = -1;
                for (int k = j; k <= V; k += v[i]) {
                    if (head <= tail && (k - queue[head]) / v[i] > s[i]) head++;
                    if (head <= tail) dp[k] = Math.max(dp[k], copy[queue[head]] + (k - queue[head]) / v[i] * w[i]);
                    while (head <= tail && copy[k] >= copy[queue[tail]] + (k - queue[tail]) / v[i] * w[i]) tail--;
                    queue[++tail] = k;
                }
            }
        }
        return dp[V];
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int V = sc.nextInt();
        int[] v = new int[N];
        int[] w = new int[N];
        int[] s = new int[N];
        for (int i = 0; i < N; i++) {
            v[i] = sc.nextInt();
            w[i] = sc.nextInt();
            s[i] = sc.nextInt();
        }
        System.out.println(backPack5(N, V, v, w, s));
    }
}