我正在寻找一种解决常见装箱问题的智能方法。给定一些具有一定容量的袋子（我这么称呼它们），以及占用一定空间的物品列表，任务是确定是否所有物品都可以放入袋子中；如果是这样，如何。我现在有一个详尽的 DFS 工作，但它需要......永远。我的 DFS 是迭代的，需要在每一步复制整个状态，这非常昂贵。这是我针对 4 个容量为 10 的袋子的特定问题的代码（如果您不想全部查看，此代码的真正相关部分只是 pack() 方法和 State 类）：
import java.util.ArrayList;
import java.util.Stack;
public class BagProblem {
int numBags;
int bagCapacity;
ArrayList<Item> items = new ArrayList<Item>();
public static void main(String[] args) {
BagProblem bp = new BagProblem(4, 10);
bp.pack();
}
public BagProblem(int numBags, int bagCapacity) {
this.numBags = numBags;
this.bagCapacity = bagCapacity;
items = new ArrayList<Item>();
items.add(new Item("item0", 6));
items.add(new Item("item1", 6));
items.add(new Item("item2", 6));
items.add(new Item("item5", 3));
items.add(new Item("item6", 3));
items.add(new Item("item7", 3));
items.add(new Item("item8", 2));
items.add(new Item("item9", 2));
items.add(new Item("item10", 2));
items.add(new Item("item11", 2));
items.add(new Item("item12", 2));
items.add(new Item("item13", 2));
items.add(new Item("item14", 1));
}
// find a valid way to pack and print the items in each Bag, or
// print failure
public void pack() {
Stack <State> s = new Stack<State>();
Bag[] currBags = new Bag[numBags];
for (int i = 0; i < numBags; i++) {
currBags[i] = new Bag(bagCapacity);
}
s.push(new State(currBags));
while(!s.isEmpty()) {
State currState = s.pop();
for (Item i : items) {
if (!currState.containsItem(i)) {
State newState = new State(currState.bags);
newState.numItems = currState.numItems;
if (newState.addItem(i)) {
s.push(newState);
if (newState.numItems == items.size()) {
System.out.println("success");
System.out.println(newState);
return;
}
}
}
}
}
System.out.println("failure");
}
private class State {
Bag[] bags;
int numItems;
public State(Bag[] currBags) {
bags = new Bag[numBags];
for (int i = 0; i < numBags; i++) {
bags[i] = new Bag(bagCapacity);
}
// figure out how to actually copy this
for (int j = 0; j < numBags; j++) {
Bag bagToCopy = currBags[j];
for (Item item : bagToCopy.contents) {
Item newItem = new Item(item.name, item.size);
bags[j].size = bagToCopy.size;
bags[j].contents.add(newItem);
}
}
}
public boolean addItem(Item i) {
for (Bag b : bags) {
if (b.addItem(i)) {
numItems++;
return true;
}
}
return false;
}
public boolean containsItem(Item i) {
for (Bag b : bags) {
for (Item item : b.contents) {
if (item.name.equals(i.name))
return true;
}
}
return false;
}
public String toString() {
String output = "";
for (Bag b : bags) {
for (Item j : b.contents) {
output += j.name + " ";
}
output += "\n";
}
return output;
}
}
private class Bag {
int capacity;
int size;
ArrayList<Item> contents;
public Bag(int capacity) {
this.capacity = capacity;
this.size = 0;
contents = new ArrayList<Item>();
}
public boolean addItem(Item i) {
if(size + i.size > capacity)
return false;
contents.add(i);
size += i.size;
return true;
}
public String toString() {
String output = "";
for (Item i : contents) {
output += i.name + " ";
}
return output + "\n";
}
}
private class Item {
String name;
int size;
public Item(String name, int size) {
this.name = name;
this.size = size;
}
public String toString() {
return name;
}
}
}
大约一百万年后，这确实给出了一个正确的答案（如果您尝试运行它，您可能不想真正等待那么长时间）：
success
item14 item7 item6 item5
item13 item12 item2
item11 item10 item1
item9 item8 item0
每行表示一个单独的包。我怎样才能加快速度？我知道尝试先放置最大的项目等有启发式方法，但我真正感兴趣的是获得基本的 DFS（或者我应该尝试回溯？）以减少开销；以后我会努力变得更漂亮的。
任何帮助将不胜感激。