C++分支界限算法解决旅行售货员问题代码

旅行售货员问题（Traveling Salesman Problem, TSP）是一个经典的组合优化问题，旨在找到一条最短路径，使得一个送货员能访问所有城市并返回起点一次。C++中的分支界限算法是一种通过搜索树结构来求解此类问题的方法，它结合了剪枝策略来减少搜索空间。

下面是一个简单的C++代码片段，展示了如何使用分支界限算法来解决TSP问题的基本框架：

#include <vector>
#include <limits>

struct Node {
    std::vector<int> path;
    int cost;
    bool operator<(const Node& other) const { return cost > other.cost; }
};

std::vector<int> tsp_brute_force(std::vector<int>& cities) {
    // Brute force solution for small inputs
    if (cities.size() <= 4) {
        // Calculate all permutations and return the shortest one
        return brute_force_permutations(cities);
    }

    // Branch and bound approach
    Node best_node = { {}, cities.size() * cities[0], true };
    std::vector<Node> frontier;

    frontier.push_back({ cities });
    while (!frontier.empty()) {
        Node current = frontier.back();
        frontier.pop_back();

        // If we found a shorter path, update the best solution
        if (current.cost < best_node.cost) {
            best_node = current;
        }

        // Generate children by adding an edge to the next city not in the path
        for (int i = 0; i < cities.size(); ++i) {
            if (std::find(current.path.begin(), current.path.end(), i) == current.path.end()) {
                Node child = current;
                child.path.push_back(i);
                child.cost = current.cost + distance(cities[current.path.back()], cities[i]);
                frontier.push_back(child);
            }
        }

        // Prune branches with higher costs
        for (Node& node : frontier) {
            if (node.cost > best_node.cost + std::numeric_limits<int>::max() / 2) {
                break;  // Early pruning
            }
        }
    }

    return best_node.path;
}

std::vector<int> brute_force_permutations(std::vector<int>& cities) {
    // Implement a full permutation algorithm here
}

// A simple Euclidean distance function as an example
int distance(int x1, int y1, int x2, int y2) {
    return std::sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}

注意：这个代码只是一个简化版的例子，实际应用中可能需要引入更复杂的剪枝规则、优先队列等数据结构，并可能需要处理更大规模的问题，这通常会借助于启发式算法如2-opt或Kruskal's algorithm来辅助决策。