Thursday, April 6, 2006 Match summary The only coder to solve all three problems in Division I, Burunduk1, won the match by a wide margin, scoring 1126.13 points, followed by liympanda with 812.74 points and asaveljevs with 808.46 points. Division II, on the other hand, was a close race, with 33 contestants solving all three problems. Newcomer owenlin won the match by his fast submissions, followed closely by .UncleMike, ar2rd, kcrtseng and mmyzf. The ProblemsPackingPartsUsed as: Division Two  Level One:
The problem statement describes the solution, so it is only necessary to correctly simulate it. One of the most elegant approaches is to simply loop over the boxes, and keep a pointer to the largest part packed so far. The solution may look as follows: public int pack(int[] parts, int[] boxes) { int part = 0; for (int b = 0; b < boxes.length && p < parts.length; b++) if (parts[p] <= boxes[b]) p++; return part; }OptimalQueues Used as: Division Two  Level Two: Used as: Division One  Level One:
This problem is one of those that are easy to code once you see the trick and convince yourself that it works. In this case, we can simply sort the customers by their arrival time. Then, we assign customers greedily into queues, from the customer who waited longest to the customer who waited shortest. Consider the case with k queues and service time T minutes. Since we have k queues, k customers will be served immediately. The service time is constant, so after T minutes all queues free up, and another batch of k customers will begin service. The third batch of k customers will begin service after having waited for 2 * T minutes. So, k customers will be served immediately, another k customers will have to wait T minutes, another k customers will have to wait 2 * T minutes, and so forth. If we assign the customers to batches by their arrival time, some customer K will end up waiting longest in total. Now, we can see that assigning this customer to either an earlier batch or a later batch could not possibly result in a better solution. All customers in the earlier batches have been waiting at least as long as K has before the bank opened. If we move K to an earlier batch, we will have to move some other customer L who has waited at least as long as K. This cannot improve the solution, because now L will be waiting at least as long as K previously did. If we move K to a later batch, K will have to wait longer in total, so again the solution will not improve. public int minWaitingTime(int[] clientArrivals, int tellerCount, int serviceTime) { int n = clientArrivals.length; Arrays.sort(clientArrivals); int ret = 0; for(int i=0; i<n; i++) { int cur = clientArrivals[n  i  1] + serviceTime * (1 + i / tellerCount); ret = (ret > cur) ? ret : cur; } return ret; }RandomizedQuickSort Used as: Division Two  Level Three:
With 55 correct submissions, this problem did not seem to cause much trouble to a number of Division 2 coders. The problem asks us to compute the expected running time of the randomized quicksort. The first step towards the solution is to understand how the described version of quicksort works. At each step, quicksort randomly selects an element of the list as the pivot. An important observation is that since the pivot is chosen randomly, the probability that the largest element will be chosen as the pivot is the same as the probability that the smallest element will be chosen, and also equal to the probability that any other given order statistic will be chosen. Since we have n possible outcomes all occurring equally likely, the probability of each one must be the same: 1/n. Once quicksort has chosen the pivot, it calls partition() to split the list into two parts: a part that contains all elements smaller than the pivot, and a part that contains all elements larger than the pivot. By the constraints of the problem, no two elements can have the same value, so no other element will be equal to the pivot. Then, quicksort calls itself recursively on both smaller lists. So, with probability 1/n, the smallest element of the list will be chosen as the pivot, and quicksort will recurse on lists of sizes 0 and n – 1. With probability 1/n, the second smallest element of the list will be chosen, and quicksort will recurse on lists of sizes 1 and n – 2. Simply extending this argument, the ith smallest element (1 <= i <= n) of the list will be chosen with probability 1/n, and will cause quicksort to call itself recursively on lists with sizes i – 1 and n – i. public double getExpectedTime(int listSize, int S, int a, int b) { if (listSize <= S) return 1.0 * b * listSize * listSize; if (cache[listSize] > 0) return cache[listSize]; double ret = 0; for(int i = 0; i < listSize; i++) { double t = getExpectedTime(i, S, a, b) + getExpectedTime(listSize  i  1, S, a, b) + 1.0 * a * listSize; ret += t / listSize; } return cache[listSize] = ret; }CageTheMonster Used as: Division One  Level Two:
This problem, like so many others, has a key idea that we must realize before successfully tackling it. First, we need to notice that if it is possible to contain the monster at all, four force fields must be sufficient. If we have more than four force fields, three fields must be parallel, and since the monster will be trapped between two of them, the third field is useless. But, if we try all possible placements of up to four fields, we end up with an O(n^6) algorithm which clearly will not run in time for n = 40. A smarter approach is to realize that the monster can always be contained by four force fields if we can place it so that it is not on the edge of the labyrinth. If the map does not allow us to place the monster so that it is not on the edge of the labyrinth, then we cannot contain it. So, since we can easily tell whether the monster can be captured using four force fields, our brute force solution would only need to check all sets of up to three fields. For each possible set of fields, we can find the connected components of the map. If some component that contains a starting point for the monster is enclosed inside the labyrinth, then the set of fields gives us one possible solution. The resulting O(n^5) algorithm may run in time, but it is a little tricky to implement correctly. Most coders used an approach that is both more efficient, as well as easier to code. The key realization is that we cannot improve a solution by moving a force field away from the monster. For example, consider a vertical force field that is to the right of the monster. The force field prevents the monster from entering any cell on the map in the column of the force field, or any column to the right of the force field. So, instead of thinking about the force field as blocking one column, we can think of it as blocking its column, as well as the entire section of the labyrinth to the right of it. Then, it becomes clear that shifting the force field to the right cannot possibly constrain the monster further. Once we realize that we can consider only creating force fields in rows and columns adjacent to the starting position of the monster, the problem becomes fairly straightforward. For each allowed starting position of the monster, there are four possible adjacent force fields we can create. We try all 16 possible subsets of them, and use depthfirst search or breadthfirst search to check whether the monster can escape. Since we have up to O(n^2) starting positions to check, and we take O(n^2) time to check whether the monster can escape, we end up with an O(n^4) algorithm that will easily run in time and is not too difficult to implement. public int capture(String[] labyrinth) { lab = labyrinth; h = lab.length; w = lab[0].length(); int ret = 10; for(int r = 1; r < h  1; r++) { for(int c = 1; c < w  1; c++) { for(int a = 0; a < 16; a++) { if (lab[r].charAt(c) != '^') continue; blockR1 = blockR2 = 2; blockC1 = blockC2 = 2; int cur = 0; if ((a & (1<<0)) != 0) { blockR1 = r  1; cur++; } if ((a & (1<<1)) != 0) { blockR2 = r + 1; cur++; } if ((a & (1<<2)) != 0) { blockC1 = c  1; cur++; } if ((a & (1<<3)) != 0) { blockC2 = c + 1; cur++; } vis = new boolean[55][55]; if (cur < ret && !escape(r,c)) ret = cur; } } } if (ret == 10) return 1; return ret; }DynamiteBoxes Used as: Division One  Level Three:
Unlike the first two Division I problems, this problem requires not only noticing a trick, but also careful implementation of the algorithm. A standard approach to problems similar to this one is to use dynamic programming whose parameters are height of the stack, some state representation of what the previous level of the stack looks like, and the size of the cluster so far. This would almost work except for one complication that makes this a Division I Hard problem: if we have two dynamitefilled boxes placed diagonally, they don’t touch, and we cannot tell whether they are connected at some lower level, or whether they belong to two separate clusters. Furthermore, if the cubes belong to two separate clusters, we need to know the sizes of both clusters to compute the rest of the problem. Given a stack with a 2 x 2 base, cubes at one level cannot belong to more than two clusters. So, one way to represent the state is using the following parameters: a bitmask that represents the previous layer of the stack, two numbers that representing the sizes of the one or two components we have at this level, and a boolean value that states that if in the previous layer the boxes were positioned diagonally, whether they belong to the same cluster. We iterate over the levels of the stack, and at each level compute the possible next states, as well as the transition probabilities. The computation of the transition probabilities is conceptually not too difficult, but we have to be careful not to make mistakes. lyimpanda’s solution roughly follows this approach. Alternatively, instead of using bitmasks to represent the previous layer, it is also possible to simply enumerate the 7 cases: no dynamite boxes, one dynamite box, two adjacent dynamite boxes, two diagonal dynamite boxes that are not connected in lower layers, two diagonal dynamite boxes that are connected, three dynamite boxes and four dynamite boxes. kalinov’s solution follows this approach. 
