- mdamircoder/CodeChef-Solutions Order food delivery and take out online from Dragon Chef Restaurant (10706 79 Ave, Grande Prairie, AB T8W 0G9, Canada). CodeChef was created as a platform to help programmers make it big in the world of Question: What do you do for path queries from u to v where one is not the ancestor of another? 3: Contest Hosting: 50: 4: Random Laddus: 200: 5 But I am receiving WA on other test cases except sample one. Sphere If b is an ancestor c, then the sum of all nodes on the path from b to c is just the sum of flat[entr[b]] until flat[entr[c]]. My sol: https://www.codechef.com/viewsolution/35511735, @shisuko There should be bit relaxation in constraint, instead of n=2100000 , n should be equal to 100000. This range is what entr[b] and entr[c] keeps track of. p_i is the point nearest p_c to its left that is higher up than it. Call i the “optimal previous” of c. Since no other point between i and c is higher than h_c, we know it is always possible to travel from p_i to p_c. Our tree has n+1 vertices, so each of the integers from 0 to 2n+1 is associated with a particular node, on either entry or exit. And suppose we have the mountain as 5 4 4 2 with tastiness as 6 7 4 3 respectively. We can then improve the score of that journey by first passing through p_i and then ending on p_c. if he starts travelling in the direction of decreasing, Chef cannot travel through solid mountain, i.e. However, there are some restrictions, Chef can only glide from a higher den to a strictly lower den, i.e. We can maintain point updates and range sum queries with either a segment tree or a Fenwick tree, giving \mathcal{O}(\log n) time per query. This way I got list of size (2n) .Made a segment Tree over it.From Euler entry and exit point , I was able to identify ,if the one node is ancestor of one or not.If yes , the sum of starting point of a to starting point of b in segment tree gives the total taste from a to b else -1. There is an \mathcal{O}(\log n)-per-query solution anyway that isn’t too different, so the bounds are fine. How about the path queries? © 2009 Directi Group.All Rights Reserved. At CodeChef we work hard to revive the geek in you by hosting a programming contest at the start of the month and two smaller programming challenges at the middle and end of the month. TLE because what if the heights are 4,5,6,7,8,9…so in worst case you are doing n jumps, we need less complexity than that. Below are the possible results: Accepted Your program ran successfully and gave a correct answer. Maintain updates on both, and use the appropriate one when querying. contests. Invitation to CodeChef June Cook-Off 2020. cook-off, cook119. Can you help me with this? days long monthly coding contest and the shorter format Cook-off and Lunchtime coding [Codechef | July long challenge 2020 | Chef and Dragon Dens | DRGNDEN | Solution] Receive points, and move Take part Editorial of Codechef july long challenge 2020 Video Editorial|video solution Beginer friendly Editorial 1)Chef and Strings : CHEFSTR1 2)Chef and Card Game : CRDGAME 3)Ada King : ADAKING 4)Missing a Point : PTMSSNG 5)Chefina and Swaps : CHFNSWPS 6)Doctor Chef : DRCHEF 7)Chef and Dragon Dens … We maintain a global timer which starts at 0, and maintain two arrays entr and ext, corresponding to the entry and exit time of each node. middle and CodeChef - A Platform for Aspiring Programmers. When doing a point update on u, make sure to update both flat[entr[u]] and flat[ext[u]]. For instance, take this coding problem move-zeroes-to-end.js. CodeChef - A Platform for Aspiring Programmers. Who are they? contests have prizes worth up to INR 20,000 (for Indian Community), $700 (for if he glides from a den at p_i to a den at p_j , then h_i \gt h_j must hold. programming I want to know where is my sol for Chef and Dragon Den going on plz? Read our Privacy Policy and Terms to know more. algorithms, computer programming, and programming Here is where you can show off your computer programming skills. CodeChef is a competitive programming community, CodeChef uses SPOJ © by Sphere That’s equivalent to c being a descendant of b, i.e. size and the likes. of Since all values are positive, a journey from b to c can always be improved if we can squeeze in another point to be visited. 2. c exists in the subtree of b. Community) and lots more CodeChef goodies up for grabs. your Warning. Use our practice section to better prepare yourself for the multiple Determining the optimal previous of all points can be done in \mathcal{O}(N) using a stack. 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