2019独角兽企业重金招聘Python工程师标准>>>

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<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 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1000H = 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;
}

#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1000H = 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;
}