Name hiding by using declaration

#include <iostream>

struct H
{
    void swap(H &rhs); 
};    

void swap(H &, H &)
{
    std::cout << "swap(H &t1, H &t2)" << std::endl;
}

void H::swap(H &rhs)
{
    using std::swap;
    swap(*this, rhs);
}


int main(void)
{
    H a;
    H b;

    a.swap(b);
}

And this is the result:

swap(H &t1, H &t2)

In the code above, I try to define a swap function of H. In the function void H::swap(H &rhs), I use an using declaration to make the name std::swap visible. If there isn't an using declaration, the code cannot be compiled because there is no usable swap function with two parameters in class H.

I have a question here. In my opinion, after I used the using declaration -- using std::swap, it just make the std::swap -- the template function in STL visible. So I thought that the swap in STL should be invoked in H::swap(). But the result showed that the void swap(H &t1, H &t2) was invoked instead.

So here is my question:

1. Why can't I invoke swap without a using declaration？(I guess it is because there is no swap function with two parameters in the class. But I am not sure. )
2. Why will the swap of my definition be invoked instead of the STL swap in the H::swap?