Thanks for your kindly consideration :)

Assume that we have an array F such that F[mask] is the number of submissions which fail only on a subset of the tests in mask. We can try all possible permutations using DP in O(2ⁿ × n) (similar to the travelling salesman DP solution). When we append a test to the current solution in DP, the number of submissions that didn't fail yet and will run on the new test is F[curMask], so we should add it to the answer of the current state. The mask in our state contains the tests which are not included yet.

Check this blog SOS Dynamic Programming [Tutorial] by usaxena95 if you need help in building the array F.

Greedy algorithm: KAN chooses the first to unlock the groups by priority of earliest second time first. This works because it unlocks something for the other to do while he unlocks other groups (if it doesn't, any other choice also doesn't).

You can always get the minimum answer if the balance between in — out degrees is 0 in every vertex. Also, if it isn't, there's no answer.

You can solve by analysing different cases.

By Dinics you meant to say that observe answer is always <5 and then create a flow problem which can have flow atmost 5 or something like this and then solve by any flow solution??

Here's a not-so-smart brute force solution.

There are at most 4 optimal places to put a block, namely the closest empty cells on the left/right of S/E. You can then iterate through no more than 2^4 different scenarios (by deciding whether or not to put a block on a candidate block) and check whether S can't reach E.

The check function can be implemented with O(n) complexity.

Solved! I made a very stupid mistake.

Start from highest value to smallest. Assign color = mex of neighbours.

dont know proof :P

Another solution that is probably less interesting but it's good to know these facts about the constructed graph:

The graph is planar, triangulated, and all vertices are boundary vertices. Check the following image:

Picking an edge and coloring its ends will split the uncolored vertices into some components, each component will have one vertex with one option, coloring that vertex will leave another vertex with one option... So after picking an edge everything will be uniquely determined unless there's a bridge, in that case you will have more than one option.

Implementation: always pick the vertex with the minimum number of remaining options and color it with the minimum available color index (just to make it work on bipartite graphs too). This solution works in O(n) since each vertex will have at most 3 options (no need for a heap or sorting).

E:

Find three nearest vertices which can propagate to each vertex(based on distance and index as mentioned in question) and then we count all those which has a red as nearest. Similarly we count number of vertices for each blue such that if we remove it will become red also for all pairs of blue similarly. Now we iterate over all pairs of blues and find the most optimal to remove by adding first counts of each blue and pair count. We need to notice that only n pairs of blue are good. rest always contribute zero to second count. So this can be done in O(n) .

I think first part of finding three vertices can be done with priority queue similar to dijkstra,

It can be solved by finding the Smallest Enclosing Circle of all visited points starting from (0, 0).

You will get a center point C and all you need now is to move the starting point so C becomes (0, 0), thus the new starting point is -C.

The algorithm is pretty much similar to your solution. I can't open it, btw :1.