Bezier curves of grade n
Just a few comments:

std::vector&lt;Vector2f&gt; make_bezier(const std::vector&lt;Vector2f&gt;&amp; anchors, double accuracy=10000.0)


You're using float for all your other floating point variables. Unless there's some really good reason to use double here (for accuracy) I'd consider changing it to a float as well.

std::vector&lt;Vector2f&gt; end;
end.push_back(anchors[0]);

for(float i=0.f; i&lt;1.f; i+=1.f/accuracy)


I'd write this a bit differently. As it stands, i can accumulate rounding errors from one iteration of the loop to the next. I'd prefer to do something like:

for (int j=0; j&lt;accuracy; j++) {
    float i = 1.0f/j;
    // ...


This way, each value of i is computed independently, and rounding from one iteration doesn't affect its value in the next.

    std::vector&lt;Vector2f&gt; temp;
    for(unsigned int j=1; j&lt;anchors.size(); ++j)
        temp.push_back(Vector2f(interpolate(anchors[j-1].x, anchors[j].x, i),
                                interpolate(anchors[j-1].y, anchors[j].y, i)));


Literally all your calls to interolate (at least in this code) happen in pairs. That being the case, I think I'd rewrite it a bit to take a couple of vector2fs as parameters, and return a vector2f containing the interpolation on both the x and y components of the inputs. It also uses i as a third input. My immediate reaction would be to define interpolate as a lambda that captured i, and took the other two parameters as inputs.

vector2f interpolate(vector2f in1, vector2f const &amp;in2) { 
    in1.x = /* ... */;
    in1.y = /* ... */;
    return in1;
}


With that in place, the loop above can be reduced to a fairly simple application of std::transform, something like this:

 std::transform(anchors.begin(), anchors.end(), 
                anchors.begin()+1, 
                std::back_inserter(temp));


Of course, pretty much the same would happen here:

        for(unsigned int j=1; j&lt;temp.size(); ++j)
            temp2.push_back(Vector2f(interpolate(temp[j-1].x, temp[j].x, i),
                                     interpolate(temp[j-1].y, temp[j].y, i)));


...as well.