贪心算法应用：多重背包启发式问题详解-海口c网

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贪心算法应用：多重背包启发式问题详解

多重背包问题是经典的组合优化问题，也是贪心算法的重要应用场景。本文将全面深入地探讨Java中如何利用贪心算法解决多重背包问题。

多重背包问题定义

**多重背包问题(Multiple Knapsack Problem)**是背包问题的变种，描述如下：

给定一个容量为W的背包
有n种物品，每种物品i有：
- 重量w_i
- 价值v_i
- 最大可用数量c_i（每种物品可以选择0到c_i个）
目标：在不超过背包容量的前提下，选择物品使总价值最大

与0-1背包（每种物品只能选0或1个）和完全背包（每种物品无限）不同，多重背包中物品有数量限制。

问题分析

数学表达

最大化：Σ(v_i * x_i) 对于i=1到n
约束条件：

Σ(w_i * x_i) ≤ W 对于i=1到n
0 ≤ x_i ≤ c_i 且 x_i为整数

关键特性

组合爆炸：可能的组合数量为Π(c_i+1)，直接枚举不可行
NP难问题：没有已知的多项式时间解法
贪心可行：虽然不能保证最优解，但能得到近似解

贪心算法解决方案

1. 基本贪心策略

三种常见的贪心策略：

价值优先：按价值降序选择
重量优先：按重量升序选择
价值密度优先：按价值/重量(v_i/w_i)降序选择

实践证明，价值密度优先策略通常效果最好。

2. Java实现基础版

import java.util.Arrays;
import java.util.Comparator;class Item {int weight;int value;int count;public Item(int weight, int value, int count) {this.weight = weight;this.value = value;this.count = count;}
}public class MultipleKnapsack {public static int greedySolution(Item[] items, int capacity) {// 按价值密度排序Arrays.sort(items, new Comparator<Item>() {@Overridepublic int compare(Item a, Item b) {double densityA = (double)a.value / a.weight;double densityB = (double)b.value / b.weight;return Double.compare(densityB, densityA); // 降序}});int totalValue = 0;int remainingCapacity = capacity;for (Item item : items) {if (remainingCapacity <= 0) break;// 计算可以取多少个当前物品int maxTake = Math.min(item.count, remainingCapacity / item.weight);if (maxTake > 0) {totalValue += maxTake * item.value;remainingCapacity -= maxTake * item.weight;}}return totalValue;}public static void main(String[] args) {Item[] items = {new Item(2, 10, 3),new Item(3, 5, 2),new Item(5, 15, 2),new Item(7, 7, 3),new Item(1, 6, 4)};int capacity = 15;System.out.println("最大价值: " + greedySolution(items, capacity));}
}

3. 算法复杂度分析

排序：O(n log n)
贪心选择：O(n)
总时间复杂度：O(n log n)

空间复杂度：O(1)（不包括输入存储）

改进的贪心算法

基础贪心算法可能不是最优的，我们可以通过以下方法改进：

1. 贪心+动态规划混合

public static int hybridSolution(Item[] items, int capacity) {// 先按贪心算法得到一个解int greedyValue = greedySolution(items, capacity);// 对高价值物品尝试动态规划Arrays.sort(items, (a, b) -> b.value - a.value);int n = items.length;int[] dp = new int[capacity + 1];for (int i = 0; i < n; i++) {Item item = items[i];for (int k = 1; k <= item.count; k++) {for (int w = capacity; w >= k * item.weight; w--) {dp[w] = Math.max(dp[w], dp[w - k * item.weight] + k * item.value);}}}return Math.max(greedyValue, dp[capacity]);
}

2. 多阶段贪心选择

public static int multiPhaseGreedy(Item[] items, int capacity) {// 阶段1：价值密度优先int phase1 = greedySolution(items, capacity);// 阶段2：纯价值优先Arrays.sort(items, (a, b) -> b.value - a.value);int phase2 = greedySolution(items, capacity);// 阶段3：纯重量优先Arrays.sort(items, (a, b) -> a.weight - b.weight);int phase3 = greedySolution(items, capacity);return Math.max(phase1, Math.max(phase2, phase3));
}

完全解与贪心解对比

动态规划解法（最优解）

public static int dpSolution(Item[] items, int capacity) {int[] dp = new int[capacity + 1];for (Item item : items) {for (int w = capacity; w >= 0; w--) {for (int k = 1; k <= item.count; k++) {if (w >= k * item.weight) {dp[w] = Math.max(dp[w], dp[w - k * item.weight] + k * item.value);}}}}return dp[capacity];
}

对比分析

方法	时间复杂度	空间复杂度	解的质量
纯贪心	O(n log n)	O(1)	近似
动态规划	O(nWC_max)	O(W)	最优
贪心+动态规划混合	O(n log n + nWC_limited)	O(W)	较好

其中C_max是最大物品数量，C_limited是限制的高价值物品数量

性能优化技巧

物品预处理：
- 移除重量>W的物品
- 合并相同物品
- 对价值密度相同的物品进行捆绑

搜索剪枝：

public static int greedyWithPruning(Item[] items, int capacity) {Arrays.sort(items, (a, b) -> Double.compare((double)b.value/b.weight, (double)a.value/a.weight));int[] best = {0};backtrack(items, 0, capacity, 0, best);return best[0];
}private static void backtrack(Item[] items, int index, int remaining, int currentValue, int[] best) {if (index == items.length || remaining == 0) {if (currentValue > best[0]) best[0] = currentValue;return;}Item item = items[index];int maxTake = Math.min(item.count, remaining / item.weight);// 从最多开始尝试，贪心顺序for (int k = maxTake; k >= 0 && (maxTake - k) <= 100; k--) {if (remaining - k * item.weight >= 0) {// 剪枝：如果剩余容量全部用下一个物品也无法超越当前最优double upperBound = currentValue + k * item.value;if (index + 1 < items.length) {upperBound += (remaining - k * item.weight) * ((double)items[index+1].value/items[index+1].weight);}if (upperBound <= best[0]) break;backtrack(items, index + 1, remaining - k * item.weight, currentValue + k * item.value, best);}}
}

并行处理：

public static int parallelGreedy(Item[] items, int capacity) {// 多种贪心策略并行执行int[] results = new int[3];Thread t1 = new Thread(() -> {Arrays.sort(items, (a, b) -> Double.compare((double)b.value/b.weight, (double)a.value/a.weight));results[0] = greedySolution(items, capacity);});Thread t2 = new Thread(() -> {Arrays.sort(items, (a, b) -> b.value - a.value);results[1] = greedySolution(items, capacity);});Thread t3 = new Thread(() -> {Arrays.sort(items, (a, b) -> a.weight - b.weight);results[2] = greedySolution(items, capacity);});t1.start(); t2.start(); t3.start();try { t1.join(); t2.join(); t3.join(); } catch (InterruptedException e) {}return Math.max(results[0], Math.max(results[1], results[2]));
}

实际应用场景

多重背包问题在现实中有广泛应用：

资源分配：服务器资源分配，选择最有价值的任务组合
投资组合：在有限资金下选择不同数量的多种投资产品
生产计划：原材料切割，最大化利用原材料
广告投放：在有限广告位中选择不同频次的广告组合
运输装载：货车装载多种货物，考虑数量限制

变种问题与扩展

1. 多维多重背包

每种物品有多个维度的重量限制（如体积和重量）：

class MultiDimensionalItem {int[] weights; // 各维度的重量int value;int count;
}public static int multiDimensionalGreedy(MultiDimensionalItem[] items, int[] capacities) {// 按综合价值密度排序Arrays.sort(items, (a, b) -> {double densityA = a.value / Arrays.stream(a.weights).sum();double densityB = b.value / Arrays.stream(b.weights).sum();return Double.compare(densityB, densityA);});int[] remaining = Arrays.copyOf(capacities, capacities.length);int totalValue = 0;for (MultiDimensionalItem item : items) {boolean canTakeMore = true;while (canTakeMore) {// 检查是否可以再取一个当前物品for (int i = 0; i < remaining.length; i++) {if (remaining[i] < item.weights[i]) {canTakeMore = false;break;}}if (canTakeMore && item.count > 0) {totalValue += item.value;item.count--;for (int i = 0; i < remaining.length; i++) {remaining[i] -= item.weights[i];}} else {break;}}}return totalValue;
}

2. 分组多重背包

物品分为若干组，每组只能选择一定数量的物品：

class Group {Item[] items;int maxSelect; // 该组最多选择的物品数
}public static int groupKnapsack(Group[] groups, int capacity) {// 两层贪心：先对组排序，再对组内物品排序Arrays.sort(groups, (a, b) -> {double maxDensityA = Arrays.stream(a.items).mapToDouble(item -> (double)item.value/item.weight).max().orElse(0);double maxDensityB = Arrays.stream(b.items).mapToDouble(item -> (double)item.value/item.weight).max().orElse(0);return Double.compare(maxDensityB, maxDensityA);});int remaining = capacity;int totalValue = 0;for (Group group : groups) {Arrays.sort(group.items, (a, b) -> Double.compare((double)b.value/b.weight, (double)a.value/a.weight));int groupSelected = 0;for (Item item : group.items) {if (groupSelected >= group.maxSelect) break;if (remaining <= 0) break;int maxTake = Math.min(item.count, remaining / item.weight);maxTake = Math.min(maxTake, group.maxSelect - groupSelected);if (maxTake > 0) {totalValue += maxTake * item.value;remaining -= maxTake * item.weight;groupSelected += maxTake;}}}return totalValue;
}

测试与验证

测试用例设计

应包含以下类型测试用例：

基础用例：简单验证功能
边界用例：空输入、单个物品、容量为0等
性能用例：大量物品测试算法效率
极端用例：所有物品重量相同、价值相同等特殊情况

JUnit测试示例

import org.junit.Test;
import static org.junit.Assert.*;public class MultipleKnapsackTest {@Testpublic void testEmptyItems() {Item[] items = {};assertEquals(0, MultipleKnapsack.greedySolution(items, 10));}@Testpublic void testZeroCapacity() {Item[] items = {new Item(2, 10, 3),new Item(3, 5, 2)};assertEquals(0, MultipleKnapsack.greedySolution(items, 0));}@Testpublic void testSingleItemWithinCapacity() {Item[] items = {new Item(5, 20, 2)};assertEquals(40, MultipleKnapsack.greedySolution(items, 10));}@Testpublic void testSingleItemExceedCapacity() {Item[] items = {new Item(15, 30, 3)};assertEquals(0, MultipleKnapsack.greedySolution(items, 10));}@Testpublic void testMixedItems1() {Item[] items = {new Item(2, 10, 3),new Item(3, 5, 2),new Item(5, 15, 2),new Item(7, 7, 3),new Item(1, 6, 4)};// 贪心解：3个重量1价值6 + 3个重量2价值10 = 3*6 + 3*10 = 48assertEquals(48, MultipleKnapsack.greedySolution(items, 15));}@Testpublic void testMixedItems2() {Item[] items = {new Item(10, 60, 1),new Item(20, 100, 1),new Item(30, 120, 1)};// 最优解是取重量20和30的物品assertEquals(220, MultipleKnapsack.hybridSolution(items, 50));}@Testpublic void testLargeInput() {// 生成100个随机物品测试性能Item[] items = new Item[100];for (int i = 0; i < 100; i++) {int w = 1 + (int)(Math.random() * 10);int v = 1 + (int)(Math.random() * 20);int c = 1 + (int)(Math.random() * 5);items[i] = new Item(w, v, c);}int capacity = 100;long start = System.nanoTime();int result = MultipleKnapsack.greedySolution(items, capacity);long end = System.nanoTime();System.out.println("Large test result: " + result);System.out.println("Time taken: " + (end - start)/1e6 + " ms");assertTrue(result > 0);}
}