We read the posts in order (ending with the first post again), and as we do, we traverse every point along the fence, marking each point in the array with $0, 1, 2, \dots$. We also record the perimeter $\ell$ (which is the number after the one we ended on). Since the fence visits each point at most once, the total time taken for this is $O(L^2)$.

Now, to compute the answer for each cow, we look up the value of both points in the array. Then, the distance along the fence in one direction is equal to the (absolute) difference between these two values; call this distance $d$. The distance in the other direction is then just $\ell - d$, so the answer for this cow is just $\min(d, \ell-d)$.

For each cow, the answer takes $O(1)$ time to compute, so the total runtime of this algorithm is $O(N+L^2)$.

C++ Code (Benjamin Qi):

N, P = map(int, input().split()) fence_posts = [tuple(map(int, input().split())) for _ in range(P)] label = [[-1] * 1001 for _ in range(1001)] perimeter = 0 def walk_segment(start, end): # walk from start to end one unit at a time global perimeter assert (start[0] == end[0]) + (start[1] == end[1]) == 1 dif = end[0] - start[0], end[1] - start[1] dist = abs(dif[0]) + abs(dif[1]) dif = dif[0] // dist, dif[1] // dist for _ in range(dist): assert label[start[0]][start[1]] == -1 label[start[0]][start[1]] = perimeter perimeter += 1 start = start[0] + dif[0], start[1] + dif[1] assert start == end for i in range(P): walk_segment(fence_posts[i], fence_posts[(i + 1) % P]) for _ in range(N): x1, y1, x2, y2 = map(int, input().split()) p1 = label[x1][y1] p2 = label[x2][y2] dist = abs(p2 - p1) dist = min(dist, perimeter - dist) print(dist)