CF1941E Rudolf and k Bridges

Bernard loves visiting Rudolf, but he is always running late. The problem is that Bernard has to cross the river on a ferry. Rudolf decided to help his friend solve this problem. The river is a grid of $ n $ rows and $ m $ columns. The intersection of the $ i $ -th row and the $ j $ -th column contains the number $ a_{i,j} $ — the depth in the corresponding cell. All cells in the first and last columns correspond to the river banks, so the depth for them is $ 0 $ .  The river may look like this.Rudolf can choose the row $ (i,1), (i,2), \ldots, (i,m) $ and build a bridge over it. In each cell of the row, he can install a support for the bridge. The cost of installing a support in the cell $ (i,j) $ is $ a_{i,j}+1 $ . Supports must be installed so that the following conditions are met: 1. A support must be installed in cell $ (i,1) $ ; 2. A support must be installed in cell $ (i,m) $ ; 3. The distance between any pair of adjacent supports must be no more than $ d $ . The distance between supports $ (i, j_1) $ and $ (i, j_2) $ is $ |j_1 - j_2| - 1 $ . Building just one bridge is boring. Therefore, Rudolf decided to build $ k $ bridges on consecutive rows of the river, that is, to choose some $ i $ ( $ 1 \le i \le n - k + 1 $ ) and independently build a bridge on each of the rows $ i, i + 1, \ldots, i + k - 1 $ . Help Rudolf minimize the total cost of installing supports.

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In the first test case, it is most profitable to build a bridge on the second row.  It is not a top view, but side view: gray cells — bridge itself, white cells are empty, black cells — supports, blue cells — water, brown cells — river bottom.In the second test case, it is most profitable to build bridges on the second and third rows. The supports will be placed in cells $ (2, 3) $ , $ (3, 2) $ , and on the river banks. In the third test case the supports can be placed along the river banks.