优先队列分支限界法解0-1背包问题

0-1背包问题都非常熟悉了，下面直接给出优先队列分支限界法解0-1背包问题的代码：

#include 
using namespace std;
class Object
{
public:
    int id;
    int weight;
    int price;
    float d;
};
class MaxHeapQNode
{
public:
    MaxHeapQNode *parent;
    int lchild;
    int upprofit;
    int profit;
    int weight;
    int lev;
};
struct cmp
{
    bool operator()(MaxHeapQNode *&a, MaxHeapQNode *&b) const
    {
        return a->upprofit < b->upprofit;
    }
};
bool compare(const Object &a, const Object &b)
{
    return a.d >= b.d;
}
int n;
int c;
int cw;
int cp;
int bestp;
Object obj[100];
int bestx[100];
void InPut()
{
    scanf("%d %d", &n, &c);
    for(int i = 1; i <= n; ++i)
    {
     scanf("%d %d", &obj[i].price, &obj[i].weight);
     obj[i].id = i;
     obj[i].d = 1.0 * obj[i].price / obj[i].weight;
    }

    sort(obj + 1, obj + n + 1, compare);
//    for(int i = 1; i <= n; ++i)
//        cout << obj[i].d << " ";
//    cout << endl << "InPut Complete" << endl;
}
int Bound(int i)
{
    int tmp_cleft = c - cw;
    int tmp_cp = cp;
    while(tmp_cleft >= obj[i].weight && i <= n)
    {
        tmp_cleft -= obj[i].weight;
        tmp_cp += obj[i].price;
        i++;
    }
    if(i <= n)
    {
        tmp_cp += tmp_cleft * obj[i].d;
    }
    return tmp_cp;
}
void AddAliveNode(priority_queue, cmp> &q, MaxHeapQNode *E, int up, int wt, int curp, int i, int ch)
{
    MaxHeapQNode *p = new MaxHeapQNode;
    p->parent = E;
    p->lchild = ch;
    p->weight = wt;
    p->upprofit = up;
    p->profit = curp;
    p->lev = i + 1;
    q.push(p);
//    cout << "加入点的信息为 " << endl;
//    cout << "p->lev = " << p->lev << " p->upprofit = " << p->upprofit << " p->weight =  " << p->weight << " p->profit =  " << p->profit << endl;
}
void MaxKnapsack()
{
    priority_queue, cmp > q; // 大顶堆
    MaxHeapQNode *E = NULL;
    cw = cp = bestp = 0;
    int i = 1;
    int up = Bound(1); //Bound(i)函数计算的是i还未处理时候的上限值
    while(i != n + 1)
    {
        int wt = cw + obj[i].weight;
        if(wt <= c)
        {
            if(bestp < cp + obj[i].price)
                bestp = cp + obj[i].price;
            AddAliveNode(q, E, up, cw + obj[i].weight, cp + obj[i].price, i, 1);
        }
        up = Bound(i + 1); //注意这里 up != up - obj[i].price而且 up >= up - obj[i].price
        if(up >= bestp) //注意这里必须是大于等于
        {
            AddAliveNode(q, E, up, cw, cp, i, 0);
        }
        E = q.top();
        q.pop();
        cw = E->weight;
        cp = E->profit;
        up = E->upprofit;
        i = E->lev;
    }
    for(int j = n; j > 0; --j)
    {
        bestx[obj[E->lev - 1].id] = E->lchild;
        E = E->parent;
    }
}
void OutPut()
{
    printf("最优装入量为 %d\n", bestp);
    printf("装入的物品为 \n");
    for(int i = 1; i <= n; ++i)
        if(bestx[i] == 1)
          printf("%d ", i);
}
int main()
{
    InPut();
    MaxKnapsack();
    OutPut();
}

测试样例：

输入：

4 7
9 3
10 5
7 2

4 1

输出：

最优装入量为 20
装入的物品为

1 3 4

运行截图：