Lucky for you, it is!

"Educational rounds" are a GREAT way to increase your rating!

I wish you high rating in every upcoming educational contest!

for E
Consider binary search on answer
Find leftmost line which will have atleast 1 unvisited cell , similarly find rightmost , topmost line and bottom-most line
This is valid if (right — left) <= 2 * x and (bottom — top) <= 2 * x if we are checking for x.

for F If N > M just transpose it
So now N ≤ 15 , so just do dp with bitmask

When you get new speed of car, you cant only check last speed limit. You need to have stack with speed limits. Example you have signs with speed limits of 100,120,150 respectively. Then , you drive 160 and you only check last speed limit which is 150. But you also must care about signs 100 120. So , maintain a stack of speeds, go back while speed.top() < curr_speed and increase counter. You can see my solution at my profile.

The problem was also featured in the Zagreb U Selection contest.

It was also asked once in the Bayan Finals.

You only need to remove (or "didn't notice") two road signs to drive without violation

Because you are passing two signs with maximum 50 with speed 100.

If there is a cycle in the graph with cost C, for every path of cost x, there is one with cost x^C.
So you can check out all the costs of simple cycles (get a spanning tree and check for each edge) and the shortest paths will be x^C1^C2... where x is the cost of any route you can find.
To find the minimum of this value we can translate the problem to linear algebra, the XOR is the same as the addition in Z/2 space, so if we treat these costs as a vector, then we can find a linearly independent set which generate these vectors in the Z^MXBITS/2 linear space using gaussian elimination.
Now we can simply get any distance from 1 to n, and XOR with the elements of the basis greedily from the most significant bit to the least.

How can he hack so fast? Just because he is on the leader board of HackerRank?(joking)

When you have more successful hacks than penalty!

in case if t==3 you might be changing previously increased speed in query 1 to -inf and than you pushed the speed limit in stack;

Try this test case-
5
1 80
3 120
3 130
3 140
1 150

No one. You drive with speed of 100. You see a sign with limit 50, you dont give a shit, so res++ You see one more sign with limit 50, you dont give a shit again, res++

Totally, you must lie about 2 signs, because you didnt slow down TWICE, not ONCE. Result is therefore 2.

If you eliminate one of the speed signals you are left with 4 1 100 3 50 1 50 which is still a violation of the rules (when you cross the 50 limit you are travelling at 100). You need 2.

Or maybe I'm too careless.....

sort(SpeedLimit.begin(), SpeedLimit.end());

I ask same question....Like LINEAR ALGEBRA WTF?! Olympiad math does not need it but Competitive programming yes..why

724G - Xor-matic Number of the Graph features the same idea.

Cant hide forever :/

If he doesn't say that, then the last overtake would violate the rules (he overtook another racer when overtakes weren't allowed)

3 1 5 2 3 5 6

It's not enough sometimes to delete the previous sign. For e. g. if there is a speed limit sign of 50, then a speed limit sign of 30, and after that he goes with 100 speed, then in your code he only has to say he didn't see the 30 sign, but then he should think the speed limit is 50, so he is still speeding.

Also you don't reset the speed limit after deleting a sign, so if there is a 30 sign, he goes with 50, then he goes with 70, then you add 2 to ans instead of 1.