01背包分支限界

算法描述：

　　活结点优先队列中结点元素N的优先级由该结点的上界函数Bound计算出的值uprofit给出。

　　子集树中以结点N为根的子树中任一结点的价值不超过N.profit。

　　可用一个最大堆来实现或节点优先队列。

　　N.weight 是结点N所相应的重量，N.profit是N所相应的价值，N.uprofit是结点N的价值上界，最大堆以这个值作为优先级。

class Object{
    friend int Knapsack(int *,int *,int ,int ,int *);
public:
    int operator <= (Object a) const
    {
        return (d >= a.d);
    }
private:
    int ID;
    float d;
};

template <class Typew,class Typep> 
class Knap;
class bbnode{
    friend Knap<int,int>;
    friend int Knapsack(int *,int *,int,int,int *);
private:
    bbnode * parent;
    bool LChild;
};

template <class Typew,class Typep>
class HeapNode{
    friend Knap<Typew,Typep>;
public:
    operator Typep() const 
    {
        return uprofit;
    }
private:
    Typep uprofit,//结点价值上界
        profit;
    Typew weight;
    int level;//活结点所相应的重量
    bbnode * ptr;
};

上界计算函数：

template <class Typew,class Typep>
class Knap<Typew,Typep>::Bound(int i)
{
    Typew cleft = c- cw;//剩余容量
    Typep b = cp;//价值上界
    //以物品单位重量价值递减序装填剩余容量
    while(i<=n && w[i] <= cleft)
    {
        cleft -= w[i];
        b += p[i];
        i++;
    }
    //装填剩余容量装满背包
    if(i<=n)
        b += p[i]/w[i] * cleft;
    return b;
}

分支限界搜索函数：

template <class Typew,class Typep>
Typep Knap<Typew,Typep>::MaxKnapsack()
{
    //优先队列式分支限界法，返回最大价值，bestx返回最优解
    //定义最大堆的容量为1000
    H = new MaxHeap< HeapNode<Typew,Typep> >(1000);
    //为bestx分配存储空间
    bestx = new int [n+1];
    //初始化
    int i=1;
    E= 0;
    cw = cp = 0;
    Typep bestp = 0;
    Typep up = Bound(1);
    //搜索子集空间树
    while(i!=n+1)//非叶节点
    {
        //检查当前扩展结点的左儿子结点
        Typew wt = cw + w[i];
        if(wt <= c)
        {
            if(cp+p[i] > bestp)
                bestp = cp+p[i];
            AddLiveNode(up,cp+p[i],cw+w[i],true,i+1);
        }
        up = Bound(i+1);

        //检查扩展结点的右儿子结点
        if(up >= bestp)//右子树 可能含有最优解
            AddLiveNode(up,cp,cw,false,i+1);
        
        //取得下一扩展点
        HeapNode<Typep,Typew> N;
        H->DeleteMax(N);
        E = N.ptr;
        cw = N.weight;
        cp = N.profit;
        up = N.uprofit;
        i = N.level;
    }
    //构造当前最优解
    for(int j=n;j>0;j--)
    {
        bestx[j] = E->LChild;
        E = E->parent;
    }
    return cp;
}

主要程序代码：

测试中.....（暂时不好使）

#include <iostream>
#include <algorithm>

class Object{
    friend int Knapsack(int *,int *,int ,int ,int *);
public:
    int operator <= (Object a) const
    {
        return (d >= a.d);
    }
private:
    int ID;
    float d;
};

template <class Typew,class Typep> 
class Knap;
class bbnode{
    friend Knap<int,int>;
    friend int Knapsack(int *,int *,int,int,int *);
private:
    bbnode * parent;
    bool LChild;
};

template <class Typew,class Typep>
class HeapNode{
    friend Knap<Typew,Typep>;
public:
    operator Typep() const 
    {
        return uprofit;
    }
private:
    Typep uprofit,//结点价值上界
        profit;
    Typew weight;
    int level;//活结点所相应的重量
    bbnode * ptr;
};

template <class Typew,class Typep>
class Knap{
    friend Typep Knapsack(Typep *,Typew *,Typew ,int ,int *);
public:
    Typep MaxKnapsack();
private:
    MaxHeap<HeapNode<Typep,Typew> > * H;
    Typep Bound(int i);
    void AddLiveNode(Typep up,Typep cp,Typew cw,bool ch,int level);
    bbnode * E;
    Typew c;
    int n;
    Typew * w;
    Typep *p;
    Typew cw;
    Typep cp;
    int *bestx;
};

template <class Typew,class Typep>
class Knap<Typew,Typep>::Bound(int i)
{
    Typew cleft = c- cw;//剩余容量
    Typep b = cp;//价值上界
    //以物品单位重量价值递减序装填剩余容量
    while(i<=n && w[i] <= cleft)
    {
        cleft -= w[i];
        b += p[i];
        i++;
    }
    //装填剩余容量装满背包
    if(i<=n)
        b += p[i]/w[i] * cleft;
    return b;
}

template <class Typep,class Typew>
void Knap<Typep,Typew>::AddLiveNode(Typep up,Typep cp,Typew cw,bool ch,int lev)
{
    //将一个新的活结点插入到子集树 和 最大堆 H中
    bbnode *b = new bbnode;
    b->parent = E;
    b->LChild = ch;
    HeapNode<Typep,Typew> N;
    N.uprofit = up;
    N.profit = cp;
    N.weight = cw;
    N.level = lev;
    N.ptr = b;
    H->Insert(N);
}

template <class Typew,class Typep>
Typep Knap<Typew,Typep>::MaxKnapsack()
{
    //优先队列式分支限界法，返回最大价值，bestx返回最优解
    //定义最大堆的容量为1000
    H = new MaxHeap< HeapNode<Typew,Typep> >(1000);
    //为bestx分配存储空间
    bestx = new int [n+1];
    //初始化
    int i=1;
    E= 0;
    cw = cp = 0;
    Typep bestp = 0;
    Typep up = Bound(1);
    //搜索子集空间树
    while(i!=n+1)//非叶节点
    {
        //检查当前扩展结点的左儿子结点
        Typew wt = cw + w[i];
        if(wt <= c)
        {
            if(cp+p[i] > bestp)
                bestp = cp+p[i];
            AddLiveNode(up,cp+p[i],cw+w[i],true,i+1);
        }
        up = Bound(i+1);

        //检查扩展结点的右儿子结点
        if(up >= bestp)//右子树 可能含有最优解
            AddLiveNode(up,cp,cw,false,i+1);
        
        //取得下一扩展点
        HeapNode<Typep,Typew> N;
        H->DeleteMax(N);
        E = N.ptr;
        cw = N.weight;
        cp = N.profit;
        up = N.uprofit;
        i = N.level;
    }
    //构造当前最优解
    for(int j=n;j>0;j--)
    {
        bestx[j] = E->LChild;
        E = E->parent;
    }
    return cp;
}

template <class Typew,class Typep>
Typep Knapsack(Typep p[],Typew w[],Typew c,int n,int bestx[])
{
    Typew W = 0;
    Typep P = 0;
    //按 单位重量价值 排序
    Object * Q = new Object [n];
    for(int i=1;i<=n;i++)
    {
        Q[i-1].ID = i;
        Q[i-1].d = 1.0*p[i]/w[i];
        P += p[i];
        W += w[i];
    }

    if(W<=c)
        return P;
    Sort(Q,n);
    
    Knap<Typew,Typep> K;
    K.p = new Typep[n+1];
    K.w = new Typew[n+1];
    for(int i=1;i<=n;i++)
    {
        K.p[i] = p[Q[i-1].ID];
        K.w[i] = p[Q[i-1].ID];
    }
    K.cp = 0;
    K.cw = 0;
    K.c = c;
    K.n = n;

    Typep bestp = K.MaxKnapsack();
    for(int j=1;j<=n;j++)
    {
        bestx[Q[i-1].ID] = K.bestx[j];
        cout<<bestx[Q[i-1.ID]]<<endl;
    }
    delete [] Q;
    delete [] K.w;
    delete [] K.p;
    delete [] K.bestx;
    cout<<"最大价值为"<<bestp<<endl;
    return bestp;
}
int main()
{
    int n,m;
    int w[100],p[100],best[100];
    cout<<"请输入想要输入的物品个数 及 背包重量:"<<endl;
    cin>>n>>m;
    cout<<"请依次输入想要输入的物品重量 及 价值"<<endl;
    for(int i=0;i<n;i++)
        cin>>w[i]>>p[i];
    Knapsack(w,p,m,n,best);
    
    return 0;
}