matlab编程分支限界法求解01背包
时间: 2024-01-31 18:02:40 浏览: 98
分支限界法是一种用于求解0/1背包问题的算法。下面是使用Matlab编程实现分支限界法求解0/1背包问题的步骤:
1. 引入库
```matlab
import java.util.PriorityQueue;
```
2. 定义物品类
```matlab
classdef Item
properties
weight
value
ratio
end
methods
function obj = Item(weight, value)
obj.weight = weight;
obj.value = value;
obj.ratio = value / weight;
end
end
end
```
3. 定义分支限界法函数
```matlab
function [maxValue, bestItems] = branchAndBound(items, capacity)
n = length(items);
queue = PriorityQueue();
queue.add(struct('level', 0, 'weight', 0, 'value', 0, 'items', []));
maxValue = 0;
bestItems = [];
while ~queue.isEmpty()
node = queue.poll();
level = node.level;
weight = node.weight;
value = node.value;
selectedItems = node.items;
if level == n
if value > maxValue
maxValue = value;
bestItems = selectedItems;
end
continue;
end
% 不选择当前物品
queue.add(struct('level', level + 1, 'weight', weight, 'value', value, 'items', selectedItems));
% 选择当前物品
if weight + items(level + 1).weight <= capacity
queue.add(struct('level', level + 1, 'weight', weight + items(level + 1).weight, 'value', value + items(level + 1).value, 'items', [selectedItems, level + 1]));
end
end
end
```
4. 测试代码
```matlab
items = [Item(2, 3), Item(3, 4), Item(4, 5), Item(5, 6)];
capacity = 8;
[maxValue, bestItems] = branchAndBound(items, capacity);
disp("Max value: " + maxValue);
disp("Best items: " + bestItems);
```
以上是使用Matlab编程实现分支限界法求解0/1背包问题的步骤和代码示例。
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