您可以添加一个参数（我们称之为link），用于收集给定掩码和节点的最佳邻居。因此它将具有与dp相同的类型。
然后在tsp运行之后，您可以使用此数据结构来重建路径。
下面是你的代码和必要的补充：

static int tsp(std::vector<std::vector<int>> const& dist, 
                size_t current, uint64_t mask, 
                std::vector<std::vector<int>>& dp,
                std::vector<std::vector<int>>& link) { // new parameter
    if (mask == ((size_t)1 << dist.size()) - 1)
        return dist[current][0];
    if (dp[mask][current] != -1)
        return dp[mask][current];

    int result = INT_MAX;
    for (uint64_t i = 0; i < dist.size(); ++i) {
        uint64_t val = (uint64_t)1 << i;
        if (mask & val)
            continue;

        // Pass extra argument in recursive call
        int sub = dist[current][i] + tsp(dist, i, mask | val, dp, link);
        if (sub < result) {
            result = sub;
            link[mask][current] = i; // Register which neighbor was best
        }
    }

    dp[mask][current] = result;
    return result;
}

// Function to translate `link` into a path, assuming 0 is starting node
static std::vector<int> getpath(std::vector<std::vector<int>>& link) {
    size_t current = 0;
    uint64_t mask = 0;
    std::vector<int> path;
    for (int i = 0; i < link[0].size(); i++) {
        path.push_back(current);
        mask |= 1 << current;
        current = link[mask][current];
    }
    return path;
}

int main() {
    std::vector<std::vector<int>> dist{
        {0, 10, 41, 31},
        {10, 0, 30, 19},
        {41, 30, 0, 20},
        {31, 19, 20, 0}
    };

    std::vector<std::vector<int>> dp(1 << dist.size(), std::vector<int>(dist.size(), -1));
   std::vector<std::vector<int>> link(1 << dist.size(), std::vector<int>(dist.size(), 0));

    int res  = tsp(dist, 0, 1, dp, link);
    std::vector<int> path = getpath(link);
    std::cout << "result " << res << "\n";
    for (auto i : path)
        std::cout << i << " ";
    std::cout << "\n";
}

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请注意，您的代码假设0是起始节点（请参阅基本情况），因此我在getpath中使用了相同的假设。