test
6
0 -20
-10 -50
-30 -100
-60 -120
-70 -130
-110 -140
my AC solution have answer
2
-140 -110
-50 -10
on this test, so you've fucked up twice

for me problem was that program was returning "axx" on "aaaxx"

Edited by author 18.03.2017 19:56

aaa  -> aaaa

by translating.............

A solution for Chinese readers.Much clearer.

This problem has obvious dynamic programming strategies. First, don't think in terms of squares, consider vertices, so the target point is (N + 1, M + 1).

--------- Algorithm 1 -----------
The most intuitive idea is dynamic programming in terms of matrices.

Let the state F [i, j] be the shortest path to the point (i, j)

State transition equation
F [i, j] = Min
{
F [i-1, j] +100
F [i, j-1] +100
F [i-1, j-1] +141.4213562373
}

Boundary condition F [0,0] = 0

F [N + 1, M + 1] is the result.

But for the 8M memory limit, a rolling array is used.

Time complexity is O (N * M)

--------- Algorithm 2 -----------
It can be found that if we only go straight, we have to go to (N + M) * 100 length. If you take the C hypotenuse, then you have to take the length of (C * 141.4213562373) + (N + M-C * 2) * 100. So obviously we want to make C as large as possible, that is, take more hypotenuse.

This can be transformed into a classic LIS model. That is, sort all the hypotenuses according to the coordinates, and then find the longest ascending sequence (x, y must be strictly increased). Taking such hypotenuses must be the optimal strategy. Then we can find C.

The result is (C * 141.4213562373) + (N + M-C * 2) * 100.

Vijos 1336 is actually an enlarged version of this question. For smaller K and large N, M, it can only be solved by algorithm 2.