一、问题描述
0-1背包问题可描述为：n个物体和一个背包。对物体i，其价值为value，重量为weight，背包的容量为W。如何选取物品装入背包，使背包中所装入的物品总价值最大？
二、算法设计
2.1 用到的数据结构
class Goods //定义货物数据类型
{
public:
int weight;//货物重量
int value;//货物价值
friend ostream& operator<<(ostream &os, const Goods &out);
};
class Knapsack//背包
{
private:
int capacity;//背包容量
int nGoodsNum;//物品数
vector<Goods> goods; //所有货物
int nMaxValue;//之前背包中装入的最大价值物品
int nCurrentWeight;//当前背包中装入物品的数量
int nCurrentValue;//当前背包中物品的价值
vector<bool> bestResult;//之前背包中物品最大价值时的物品
vector<bool> currentResult;//当前背包中的物品
}
2.2 算法步骤
1)定义解空间。（X0 , X1，X2，X3…..Xn）,Xi的值为true或false。
(i = 0,1,2,3….n)
2)确定解空间。问题的解空间描述了2^n种可能的解，采用一个二叉满树组织，解空
间的深度为问题的规模。
3)搜索解空间
a.约束条件。背包中物品重量小于背包容量。
b.限界条件。nCurrentValue为当前背包中物品价值，nMaxValue之前背包中装
入的最大价值的物品。nP = bound(i + 1)，第 i个物品之后的所有物品可装
入背包的最大价值。要求：nP + nCurrentValue > nMaxValue.
c.以广度优先的方式进行搜索.首先以根节点即第一个物品开始搜索.
三、算法描述
int bound(int i)//限界函数
{
int nLeftCapacity = capacity - nCurrentWeight;
int tempMaxValue = nCurrentValue;
while (i < nGoodsNum && goods[i].weight <= nLeftCapacity)
{
nLeftCapacity -= goods[i].weight;
tempMaxValue += goods[i].value;
}
if (i < nGoodsNum)
{
tempMaxValue += (float)(goods[i].value) / goods[i].weight * nLeftCapacity;
}
return tempMaxValue;
}
void knapsack_0_1_branchAndBound()
{
list<Node*> activeNodes;
list<Node*> diedNodes; sortByUintValue()
//判断第一个物品是否可放入背包
if (goods[0].weight < capacity)
{
activeNodes.push_back(new Node(nullptr, true, goods[0].weight, goods[0].value,0));
}
activeNodes.push_back(new Node(nullptr, false, 0,0,0));
Node *curNode = nullptr;
Node *preNode = nullptr;
int curId;
while (!activeNodes.empty())
{
//取出队列中最靠前的节点
curNode = activeNodes.front();
activeNodes.pop_front();
diedNodes.push_back(curNode);//放入死节点队列
if (curNode->id + 1 == nGoodsNum)//如果当前出队节点为最低层节点
{
continue;
}
if (nMaxValue < curNode->nValue)//若此节点物品价值大于背包中最大物品价值
{//将背包中物品替换为当前节点所表示物品
nMaxValue = curNode->nValue;
preNode = curNode;
while (nullptr != preNode)
{
bestResult[preNode->id] = preNode->leftChild;
preNode = preNode->parent;
}
}
nCurrentValue = curNode->nValue;
nCurrentWeight = curNode->nWeight;
curId = curNode->id;
if (nMaxValue >= bound(curId + 1))//剪枝
{
continue;
}
//判断下个物品是否可加入队列
if (nCurrentWeight + goods[curId + 1].weight <= capacity)
{
activeNodes.push_back(new Node(curNode, true, nCurrentWeight + goods[curId + 1].weight, nCurrentValue + goods[curId + 1].value, curId + 1));
}
activeNodes.push_back(new Node(curNode, false, nCurrentWeight,nCurrentValue,curId + 1));
}
while (!diedNodes.empty())
{
curNode = diedNodes.front();
delete curNode;
diedNodes.pop_front();
}
}
四、算法复杂性分析
4.1 空间复杂性：限界函数为O(1),最坏情况下需搜索2^(n +１) –２个节点，需O(2^n )个空间存储节点，则算法空间复杂性为O(2^n )。
4.2 时间复杂性：限界函数时间复杂度为O(n),而最坏情况有2^(n +１) – ２个节点，若对每个节点用限界函数判断，则其时间复杂度为O(n2^n).而算法中时间复杂度主要依赖限界函数，则算法的时间复杂度为O(n2^n)。
五、算法实现与测试
5.1 代码实现
#include <iostream>
#include <vector>
#include <stack>
#include <algorithm>
#include <list>
using namespace std;
const int NKNAPSACKCAP = 10;
class Goods //定义货物数据类型
{
public:
int weight;
int value;
friend ostream& operator<<(ostream &os, const Goods &out);
};
ostream& operator<<(ostream &os, const Goods &out)
{
os << "重量：" << out.weight << " 价值： " << out.value;
return os;
}
typedef vector<Goods> AllGoods;//定义所有货物数据类型
class Node
{
public:
Node *parent;
int nWeight;
int nValue;
int id;
bool leftChild;
Node(Node *_parent, bool _left, int weight, int value, int _id) :parent(_parent), leftChild(_left), nWeight(weight), nValue(value), id(_id)
{}
~Node()
{
}
};
class Knapsack
{
private:
int capacity;//背包容量
int nGoodsNum;//物品数
vector<Goods> goods;
int nMaxValue;
int nCurrentWeight;
int nCurrentValue;
vector<bool> bestResult;
int bound(int i)
{
int nLeftCapacity = capacity - nCurrentWeight;
int tempMaxValue = nCurrentValue;
while (i < nGoodsNum && goods[i].weight <= nLeftCapacity)
{
nLeftCapacity -= goods[i].weight;
tempMaxValue += goods[i].value;
}
if (i < nGoodsNum)
{
tempMaxValue += (float)(goods[i].value) / goods[i].weight * nLeftCapacity;
}
return tempMaxValue;
}
public:
Knapsack(AllGoods &AllGoods, int nKnapsackCap)
{
nGoodsNum = AllGoods.size();
capacity = nKnapsackCap;
nCurrentWeight = 0;
nCurrentValue = 0;
nMaxValue = 0;
for (int i = 0; i < nGoodsNum; ++i)
{
goods.push_back(AllGoods[i]);
bestResult.push_back(false);
}
}
void sortByUintValue()
{
stable_sort(goods.begin(), goods.end(), [](const Goods& left, const Goods& right)
{return (left.value * right.weight > left.weight * right.value); });
}
void printGoods()
{
for (size_t i = 0; i < goods.size(); ++i)
{
cout << goods[i] << endl;
}
}
void printResult()
{
cout << "MAX VALUE: " << nMaxValue << endl;
for (int i = 0; i < nGoodsNum; ++i)
{
if (bestResult[i])
{
cout << goods[i] << endl;
}
}
}
void knapsack_0_1_branchAndBound()
{
list<Node*> activeNodes;
list<Node*> diedNodes; sortByUintValue();
//判断第一个物品是否可放入背包
if (goods[0].weight < capacity)
{
activeNodes.push_back(new Node(nullptr, true, goods[0].weight, goods[0].value,0));
}
activeNodes.push_back(new Node(nullptr, false, 0,0,0));
Node *curNode = nullptr;
Node *preNode = nullptr;
int curId;
while (!activeNodes.empty())
{
//取出队列中最靠前的节点
curNode = activeNodes.front();
activeNodes.pop_front();
diedNodes.push_back(curNode);//放入死节点队列
if (curNode->id + 1 == nGoodsNum)//如果当前出队节点为最低层节点
{
continue;
}
if (nMaxValue < curNode->nValue)//若此节点物品价值大于背包中最大物品价值
{//将背包中物品替换为当前节点所表示物品
nMaxValue = curNode->nValue;
preNode = curNode;
while (nullptr != preNode)
{
bestResult[preNode->id] = preNode->leftChild;
preNode = preNode->parent;
}
}
nCurrentValue = curNode->nValue;
nCurrentWeight = curNode->nWeight;
curId = curNode->id;
if (nMaxValue >= bound(curId + 1))//剪枝
{
continue;
}
//判断下个物品是否可加入队列
if (nCurrentWeight + goods[curId + 1].weight <= capacity)
{
activeNodes.push_back(new Node(curNode, true, nCurrentWeight + goods[curId + 1].weight, nCurrentValue + goods[curId + 1].value, curId + 1));
}
activeNodes.push_back(new Node(curNode, false, nCurrentWeight,nCurrentValue,curId + 1));
}
while (!diedNodes.empty())
{
curNode = diedNodes.front();
delete curNode;
diedNodes.pop_front();
}
}
};
//获取物品信息，此处只是将书上例子输入allGoods
void GetAllGoods(AllGoods &allGoods)
{
Goods goods;
goods.weight = 2;
goods.value = 6;
allGoods.push_back(goods);
goods.weight = 2;
goods.value = 3;
allGoods.push_back(goods);
goods.weight = 2;
goods.value = 8;
allGoods.push_back(goods);
goods.weight = 6;
goods.value = 5;
allGoods.push_back(goods);
goods.weight = 4;
goods.value = 6;
allGoods.push_back(goods);
goods.weight = 5;
goods.value = 4;
allGoods.push_back(goods);
}
int main()
{
AllGoods allGoods;
GetAllGoods(allGoods); //要求按照单位物品价值由大到小排序
Knapsack knap(allGoods,NKNAPSACKCAP);
knap.printGoods();
knap.knapsack_0_1_branchAndBound();
knap.printResult();
return 0;
}
5.2 运行结果
注意：代码中可能会用到C++11新特性，请在支持C++11标准的环境下编译运行
(如VS2013)
