### rng_58's blog

By rng_58, history, 3 years ago, We will hold Keyence Programming Contest 2020.

The point values will be 100 — 200 — 400 — 700 — 900 — 1100.

We are looking forward to your participation!  Comments (50)
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 » I keep wanting to participate in these unrated (for me) contests, but I always oversleep or play a few too many AOEII games or just forget. Surely this time...
•  » » Forgot again?
•  » » » Kinda — I checked out F when the contest was already running, realised that the only difficult part is going to be tiebreaking in a scheme "remove monochrome rows/columns while possible", since the number of possibilities is given just by the number of used rows/columns, and decided that I don't want to spend the remaining time in the contest on this.
•  » » Atcoder supports virtual contest now.
 » Is this contest at Div.2 difficulty?
•  » » This is an "except for red" contest, so it's more like Div 1.5.
•  » » I think it is more difficult than Div1. I only solve 1 problem in the last contest...
•  » » » 3 years ago, # ^ | ← Rev. 2 →   I don't think a candidate master even can't solve a 200-point task in atcoder. Let's wait and see :D Anyway,I think it's going to be an excellent contest!
•  » » » » Wish me good luck this time!
 » is D bitmask DP??if yes, where my solution is going wrong
•  » » Your cost is wrong.
•  » » » thanks i got it
 » how to solve 2nd ans 4th question??In second question, i tried to store count of overlap for each i ,and later sort and remove maximum overlapped element ,but got WA.
•  » » Represent each robot as an interval [X-L, X+L], then use dp + binary search to find largest subset of intervals of intervals whose intersection is empty
•  » » » Bit of an overkill actually, just pick the Segment with smallest end point and then remove all intersection ones, keep doing and keep a count.
 » how to solve question 2??
•  » » Solved it just like the Interval Scheduling problem. Found the minimum number of intervals which are overlapping should be removed (using the standard greedy approach) and then the maximum robots we can keep is Total — minimum removed. https://www.geeksforgeeks.org/activity-selection-problem-greedy-algo-1/
 » AC on F 3 minutes after the contest QAQGreat round as always! I particularly liked D and F.
•  » » Can you please explain D?
•  » » » When we reorder the elements, if the card that was initially at position $i$ is at position $j$, then it is red face up if $i \equiv j\ (\text{mod}\ 2)$, and otherwise blue face up. So we only care about the final order the cards end up in, not the sequence of swaps.The optimal way to get to a particular final order is to first make swaps to move the card that should be in the first position to the first position, then do the same for the second and so on.Therefore, we can do DP with states indicating the elements we have already used and the value the current sequence ends in. From any particular DP state, loop all cards not yet used. The $d$th not yet used card costs $d$ to use as the next element in the sequence, and we know which way up it would be, so we know whether it is larger than the previous one, and what value the new sequence would end in.After calculating the DP, just output the minimum DP value where all cards are used and the sequence ends in an arbitrary value. code#include #include #include using namespace std; using ll = long long; const ll INF = (int)1e9; const int N = 18; const int V = 50; int dp[1 << N][V+1]; int as[N]; int bs[N]; int off[N][N]; void cap(int& v, int off) { if (v > off) v = off; } int main() { ios_base::sync_with_stdio(false); cin.tie(0); int n; cin >> n; for (int i = 0; i < n; ++i) cin >> as[i]; for (int i = 0; i < n; ++i) cin >> bs[i]; for (int i = 0; i < n; ++i) { for (int j = 0; j < n; ++j) { off[i][j] = (i % 2 == j % 2) ? as[j] : bs[j]; } } int h = 1 << n; for (int m = 0; m < h; ++m) { for (int j = 0; j <= V; ++j) dp[m][j] = INF; } dp = 0; for (int m = 0; m < h; ++m) { int i = __builtin_popcount(m); for (int v = 0; v <= V; ++v) { int d = 0; for (int j = 0; j < n; ++j) { if (m & (1 << j)) continue; if (off[i][j] >= v) cap(dp[m | (1 << j)][off[i][j]], dp[m][v] + d); ++d; } } } int res = INF; for (int v = 0; v <= V; ++v) res = min(res, dp[h-1][v]); cout << (res >= INF ? -1 : res) << '\n'; } 
•  » » » » There is also Greedy solution without DP.
•  » » » » » Can you please explain!
•  » » » I just use easy implement.My code is easy to understand. https://atcoder.jp/contests/keyence2020/submissions/9573589
•  » » » » its so cool
•  » » » 3 years ago, # ^ | ← Rev. 2 →   My solution slightly differs from the ones present here. It doesn't use DP.After all the swaps, there will some cards with the red face up, others with the blue face. Iterate through all possible $2^{n}$ combinations. A particular combination is valid if we get an increasing sequence after a sequence of swaps. Cards with the red face up have to move an even distance to their final position while ones with the blue face move an odd distance. Iterate the cards from the left and greedily assign them the lowest available final position. If its not possible to do so, the combination is invalid. To calculate the minimum number of swaps to make this final sequence, simply calculate the number of inversions of the final sequence w.r.t the original sequence. Repeat the above procedure for all possible $2^n$ combinations. The minimum of all of them is the answer.
 » Do you have a solution？
 » Can someone give me the data of problem E 01-02. I've been WA for so long and I can't find what's wrong.
•  » » I find that even the data of keyence 2019 haven't been put into dropbox....
•  » » » What is "dropbox"?
•  » » » » Atcoder testcases are uploaded here: https://www.dropbox.com/sh/arnpe0ef5wds8cv/AAAk_SECQ2Nc6SVGii3rHX6Fa?dl=0
 » Why was Task C so much easier than B even though its 400 points ?
•  » » I think that B's coding difficulty is much higher than C. But I think these two problem's difficulty in thinking is almost the same. (Which one is more solvable depend on yourself.)
 » I think I misinterpreted Problem B, for test case 2 1 5 10 5, shouldn't the answer be 2 as we had to exclude the end points? Can someone tell me why its 1, as we had to exclude the end points, the ranges come out to be [-3,5] and [6,14] and they aren't intersecting.
•  » » 3 years ago, # ^ | ← Rev. 3 →   I also misunderstood "excluding the endpoints". During the contest, I think the movable range of arms of robot i is [xi − li + 1, xi + li − 1], not [xi − li, xi + li − 1]
•  » » » Excluding endpoints mean, that robot can reach any point(not necessary discrete) in interval (xi — li, xi + li), but cannot reach xi — li and xi + li.
•  » » » » Ok...I see. Thanks.
•  » » » » 3 years ago, # ^ | ← Rev. 2 →   2 1 5 10 5 Why is the answer for this case 1 and not 2 by your logic? Robot 1 can reach (-4, 6) and Robot 2 can reach (5, 15) without intersecting.
•  » » » » » They intersect on interval (5, 6)
•  » » » » » » I think I do get your point, the question actually wants to say that hands can touch thats why (5,6) isn't allowed, (5,5) is because that just means that they are touching and not actually intersecting. Thanks for clearing it.
•  » » » » despair Can you please see what am I doing wrong?? code
•  » » » » » The only thing wrong with your code is the way you are sorting, you have to sort the given ranges with respect to their right end. Here's the AC code with just the sorting changed — Code
 » Could someone share their approach for E?
•  » » 3 years ago, # ^ | ← Rev. 2 →   Main ideas: If two adjacent vertices $u$ and $v$ have $D_u = D_v$, it's possible to satisfy both by putting weight $D_u$ on the edge between them and giving the vertices opposite colors. If two adjacent vertices $u$ and $v$ have $D_u < D_v$, it's possible to satisfy $v$ by putting weight $D_v - D_u$ on the edge between them and giving the vertices the same color. If a vertex $v$ has no neighbor $u$ with $D_u \leq D_v$, it's not possible to satisfy $v$. If no vertex violates (3), it's always possible to construct a valid coloring and weight assignment using (1) and (2). Implementation is straightforward if vertices are processed in non-decreasing order of $D$. Unused edges can be assigned a weight of $10^9$.
 » In problem B, why do we have to sort according to x+l and not x-l?
•  » » I sort by x-l,but implement strategy need some change.Sort by x+l I always see in some textbook. https://atcoder.jp/contests/keyence2020/submissions/9573530
 » How to solve F?
•  » » 3 years ago, # ^ | ← Rev. 3 →   We can reduce the problem to the following: Suppose that the board is initially uncolored. If exactly $i$ rows and $j$ columns are never chosen, then how many possible ways are there to color the grid? Either $i=j=0$ or $i,j>0.$A way to color the grid is valid if we can repeatedly remove rows or columns in which all of the squares are the same color until only uncolored squares are left. Call rows and columns that may be removed uniform. To count each such grid exactly once, we should specify an order in which uniform rows and columns should be removed. Step 1: Remove all uniform columns. Step 2: Remove all uniform rows. Step 3: Remove all uniform columns created by the removal of the rows in step 2. ... and so on. Let $d$ be the number of distinct colors among the uniform rows or columns removed in the $i$-th step. It can be shown that $d$ is non-increasing over time (ignoring step 1). Furthermore, the number of ways to remove rows/columns on the $i+1$-st step depends only on $d$ after the $i$-th step and the number of rows/columns remaining. For example, if only black rows were removed during step 2 ($d=1$), then no black columns can be removed during step 3, because they were already removed during step 1. This means that $d$ will still equal one for step 3 (unless no columns are removed at all). Note that when $x=y=0$, $d$ is equal to two after every step (again, ignoring step 1).Code
•  » » » 3 years ago, # ^ | ← Rev. 2 →   .
 » Is there the tutorial in English?