优先队列分支限界求解最优装载c++代码

以下是使用优先队列分支限界算法求解最优装载问题的C++代码实现：

#include <iostream>
#include <queue>
using namespace std;

struct Node {
    int level;      // 当前节点所在的层数
    int weight;     // 当前节点的重量
    int profit;     // 当前节点的价值
    float bound;    // 当前节点的价值上界
};

// 重载小于运算符，将节点按照价值上界从大到小排序
bool operator<(const Node& a, const Node& b) {
    return a.bound < b.bound;
}

// 计算一个节点的价值上界
float bound(Node u, int n, int capacity, int* w, int* p) {
    if (u.weight >= capacity) {
        return 0;
    }

    float bound = u.profit;
    int j = u.level + 1;
    int totweight = u.weight;
    while ((j < n) && (totweight + w[j] <= capacity)) {
        totweight += w[j];
        bound += p[j];
        j++;
    }

    if (j < n) {
        bound += (capacity - totweight) * p[j] / w[j];
    }

    return bound;
}

// 使用优先队列分支限界算法求解最优装载问题
int knapsack(int n, int capacity, int* w, int* p) {
    priority_queue<Node> Q;
    Node u, v;
    int* order = new int[n];

    for (int i = 0; i < n; i++) {
        order[i] = i;
    }

    // 按照单位价值从大到小排序
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
            if (1.0 * p[order[i]] / w[order[i]] < 1.0 * p[order[j]] / w[order[j]]) {
                swap(order[i], order[j]);
            }
        }
    }

    u.level = -1;
    u.profit = 0;
    u.weight = 0;

    float maxprofit = 0;
    Q.push(u);

    while (!Q.empty()) {
        u = Q.top();
        Q.pop();

        if (u.bound > maxprofit) {
            v.level = u.level + 1;
            v.weight = u.weight + w[v.level];
            v.profit = u.profit + p[v.level];

            if (v.weight <= capacity && v.profit > maxprofit) {
                maxprofit = v.profit;
            }

            v.bound = bound(v, n, capacity, w, p);

            if (v.bound > maxprofit) {
                Q.push(v);
            }

            v.weight = u.weight;
            v.profit = u.profit;
            v.bound = bound(v, n, capacity, w, p);

            if (v.bound > maxprofit) {
                Q.push(v);
            }
        }
    }

    delete[] order;
    return maxprofit;
}

int main() {
    int n = 5;
    int capacity = 10;
    int w[] = {2, 2, 6, 5, 4};
    int p[] = {6, 3, 5, 4, 6};

    int maxprofit = knapsack(n, capacity, w, p);
    cout << "The maximum profit is " << maxprofit << endl;

    return 0;
}

以上代码仅供参考，具体实现可能因问题而异。