How acceleration/deceleration values are calculated?

[ProVival]

Hi, thanks for your work! I was on the process of learning SIPP-IP and was wondering how the values here are calculated and what do they mean exactly?
vector<pair<int, int>> costs = {make_pair(0, 20), make_pair(0, 29), make_pair(20, 15), make_pair(28, 12), make_pair(34, 6)};

Edit: Now i did understand after calculating the values right :D sorry for bothering about that. But i do have couple more questions

In the code below from applyPrimitive() function, in which cases there are more than one end-cell and when is it helpful?

if (new_tlower <= new_tupper && (mv.isEndCell && !endCellTouched) &&
                    mp.v == 0)
                { // if there is at least one timestep to get there and vel there is zero and this is the first end-cell in the primitive.
                    new_tupper = itr->first - 1 - mv.swt;
                }

Also here in comparison operator, any particular reason to compare based on this? Is the number 4 equal to MXO?
((a.x * MXW + a.y) * 4 + a.o) * MXV + a.v < ((b.x * MXW + b.y) * 4 + b.o) * MXV + b.v

Finally most important part for me. For example i want to have a=1m/s^2 and v_max = 1m/s, i also added a motion primitive so that it goes for just 1 cell and then stops like the following:

tmp.mvs = {Primitive::move(0, 0, 0, 20, 0), Primitive::move(dx[o], dy[o], 0, 20, 1)};
tmp.o = o;
tmp.v = 0;
motion_primitives[o][0].emplace_back(tmp);

But i can't figure out how should i modify motion primitves for acceleration and deceleration. Any chance you can give me an example for that? With my acceleration and v_max, it will have the max speed after moving 0.5m at t=1 and then it have to move the remaining 0.5m in 0.5s. So it will take 1.5seconds = 15 timestep to sweep first cell (cell we currently in). And to sweep through second cell, it has to go 10 timestep. So i tried to add the following to my motion primitives:

// acceleration primitives
tmp.v = 1; // at the end of primitive, v=1
tmp.mvs = {Primitive::move(0, 0, 0, 15, 0), Primitive::move(dx[o], dy[o], 0, 25 1)};
motion_primitives[o][0].push_back(tmp);
tmp.mvs.clear();

But with setting second moves sweeping time as 25 i feel like the solution states didn't make sense, so i changed it to 15 but i still don't know if it's correct or not.
I am really confused on how should i approach this, i would be really grateful if you could help me. Again, thanks for your amazing work!