裝載問題實質： 裝載問題是一個子集選取問題，因此其解空間樹是一顆子集樹。

這裡實現優先佇列式分支限界法。

#include <bits/stdc++.h>
using namespace std;
class MaxHeapQNode
{
public:
    MaxHeapQNode *parent;
    int lchild;
    int weight;
    int lev;
};
struct cmp
{
    bool operator()(MaxHeapQNode *&a, MaxHeapQNode *&b) const
    {
        return a->weight < b->weight;
    }
};
int n;
int c;
int bestw;
int w[100];
int bestx[100];
void InPut()
{
    scanf("%d %d", &n, &c);
    for(int i = 1; i <= n; ++i)
        scanf("%d", &w[i]);
}
void AddAliveNode(priority_queue<MaxHeapQNode *, vector<MaxHeapQNode *>, cmp> &q, MaxHeapQNode *E,  int wt, int i, int ch)
{
    MaxHeapQNode *p = new MaxHeapQNode;
    p->parent = E;
    p->lchild = ch;
    p->weight = wt;
    p->lev = i + 1;
    q.push(p);
}
void MaxLoading()
{
    priority_queue<MaxHeapQNode *, vector<MaxHeapQNode *>, cmp > q; // 大頂堆
    //定義剩餘重量陣列r
    int r[n + 1];
    r[n] = 0;
    for(int j = n - 1; j > 0; --j)
        r[j] = r[j + 1] + w[j + 1];
    int i = 1;
    MaxHeapQNode *E;
    int Ew = 0;
    while(i != n + 1)
    {
        if(Ew + w[i] <= c)
        {
            AddAliveNode(q, E, Ew + w[i] + r[i], i, 1);
        }
        AddAliveNode(q, E, Ew + r[i], i, 0);

        //取下一節點
        E = q.top();
        q.pop();
        i = E->lev;
        Ew = E->weight - r[i - 1];
    }
    bestw = Ew;
    for(int j = n; j > 0; --j)
    {
        bestx[j] = E->lchild;
        E = E->parent;
    }
}
void OutPut()
{
    printf("最優裝載量為 %d\n", bestw);
    printf("裝載的物品為 \n");
    for(int i = 1; i <= n; ++i)
        if(bestx[i] == 1)
          printf("%d ", i);
}
int main()
{
    InPut();
    MaxLoading();
    OutPut();
}
 

測試樣例：

輸入：

(多解，輸出其中一個）

4 60
10
40

50

20

輸出

最優裝載量為 60
裝載的物品為
1 3

執行截圖