Trip Shopping

Let’s strip the game down to what it actually is.

For every index $i$, replace the pair $(a_i,b_i)$ with the segment

$$[l_i,r_i] = [\min(a_i,b_i), \max(a_i,b_i)].$$

Its contribution to the answer is

$$|a_i-b_i| = r_i-l_i.$$

So the starting value is

$$S = \sum_{i=1}^n (r_i-l_i).$$

Now we need to understand exactly what happens when Ali picks two indices.

What Bahamin can do to two indices

Suppose Ali picks indices $i$ and $j$. Bahamin gets the four numbers

$$l_i, r_i, l_j, r_j$$

and can rearrange them into two new pairs. To maximize the sum of two absolute differences, sort those four numbers:

$$x_1 \le x_2 \le x_3 \le x_4.$$

The best possible value is

$$(x_4-x_1) + (x_3-x_2).$$

That is: pair the smallest with the largest, and the two middle values together. Pairing big with small is the whole trick. Very subtle, very evil, very Codeforces.

Now compare that to the old value of just these two indices:

$$(r_i-l_i) + (r_j-l_j).$$

There are two cases.

Case 1: the two segments overlap

If $[l_i,r_i]$ and $[l_j,r_j]$ intersect, then after sorting the four endpoints, the maximized value is exactly the same as the old total.

Example shape:

$$l_i \le l_j \le r_i \le r_j.$$

Old value:

$$(r_i-l_i)+(r_j-l_j).$$

Bahamin’s max value:

$$(r_j-l_i)+(r_i-l_j).$$

These are equal after expanding. So the gain is $0$.

Case 2: the two segments are disjoint

Say

$$r_i < l_j.$$

The gap between them is

$$g = l_j-r_i.$$

Old value:

$$(r_i-l_i)+(r_j-l_j).$$

Bahamin’s max value:

$$(r_j-l_i)+(l_j-r_i).$$

Subtract old from new:

$$[(r_j-l_i)+(l_j-r_i)] - [(r_i-l_i)+(r_j-l_j)] = 2(l_j-r_i)=2g.$$

So selecting two disjoint segments with gap $g$ increases the total by exactly $2g$.

That means every possible first move has a cost to Ali:

$$\text{gain}(i,j)=2\cdot \max(0,\text{gap between segments }i,j).$$

Ali wants the smallest possible gain.

Why $k$ basically does not matter

This is the part where the problem tries to look scarier than it is.

Since $k \ge 1$, Ali must choose some pair at least once. Bahamin then maximizes that pair. After that, if Ali chooses the same pair again, Bahamin cannot make it any worse: the two selected indices are already arranged in the maximum possible way for those four numbers.

So Ali can just pick his best pair in every round. Extra rounds do not force him to touch more indices. Nice try, statement.

Therefore the final answer is

$$S + \min_{i<j} \text{gain}(i,j).$$

Equivalently:

$$S + 2 \cdot \min_{i<j} \max(0,\text{gap between segments }i,j).$$

If any two segments overlap, the minimum gap is $0$, and the answer is just $S$.

Finding the minimum gap

We have $n$ segments $[l_i,r_i]$.

Sort them by $l_i$.

While scanning from left to right, keep the largest right endpoint seen so far, call it $mx$.

For the current segment $[l,r]$:

• if $l \le mx$, it overlaps some previous segment, so the minimum gap is $0$;
• otherwise, its nearest possible previous segment gap is $l-mx$.

Why is $mx$ enough? Among all previous segments, the one with the largest right endpoint gives the smallest gap to the current left endpoint. Anything ending earlier is worse.

So the scan gives the minimum gap in $O(n)$ after sorting.

Total complexity per test case:

$$O(n\log n)$$

which is completely fine for $\sum n \le 2\cdot 10^5$.