用c++伪代码编写一个最小耗费分支限界算法寻找最小的顶点覆盖，并确定算法的复杂度。-Go服务器开发

以下是一个使用C++伪代码编写的最小耗费分支限界算法来寻找最小的顶点覆盖：

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

struct Node {
    int vertex;
    int cost;
    vector<int> selectedVertices;
};

int minVertexCover(vector<vector<int>>& graph, vector<bool>& selected) {
    int n = graph.size();
    queue<Node> q;

    Node root;
    root.vertex = 0; // 从顶点0开始
    root.cost = 0;
    root.selectedVertices.resize(n, false);
    
    q.push(root);

    int minCost = INT_MAX;

    while (!q.empty()) {
        Node current = q.front();
        q.pop();

        int level = current.vertex + 1; // 下一层要处理的顶点

        if (level == n) { // 到达叶子节点，更新最小耗费值
            if (current.cost < minCost)
                minCost = current.cost;
            continue;
        }

        Node leftChild, rightChild;

        leftChild.vertex = level;
        rightChild.vertex = level;

        leftChild.cost = current.cost; // 左子树不选择当前顶点
        rightChild.cost = current.cost + 1; // 右子树选择当前顶点

        leftChild.selectedVertices.assign(current.selectedVertices.begin(), current.selectedVertices.end());
        rightChild.selectedVertices.assign(current.selectedVertices.begin(), current.selectedVertices.end());

        q.push(leftChild);
        
        if (!selected[current.vertex]) {
            rightChild.selectedVertices[current.vertex] = true;
            q.push(rightChild);
         }
    }

    return minCost;
}

int main() {
    int n; // 顶点数量
    cin >> n;

    vector<vector<int>> graph(n, vector<int>(n));

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> graph[i][j]; // 输入邻接矩阵
        }
    }

    vector<bool> selected(n, false); // 记录被选择的顶点

    int minCoverCost = minVertexCover(graph, selected);

    cout << "Minimum vertex cover cost: " << minCoverCost << endl;

    return 0;
}

该算法的复杂度为O(2^n)，其中n为图的顶点数量。因为最小耗费分支限界算法需要对每个顶点进行选择和不选择的判断，而对于每个顶点都有两种选择，所以时间复杂度是指数级别的。

内容由零声教学AI助手提供，问题来源于学员提问