606A — Magic Spheres. Let’s count how many spheres of each type are lacking to the goal. We must do at least that many transformations. Let’s count how many spheres of each type are extra relative to the goal. Each two extra spheres give us an opportunity to do one transformation. So to find out how many transformations can be done from the given type of spheres, one must look how many extra spheres there are, divide this number by 2 and round down. Let’s sum all the opportunities of transformations from each type of spheres and all the lacks. If there are at least that many opportunities of transformations as the lacks, the answer is positive. Otherwise, it’s negative.
606B — Testing Robots. Let’s prepare a matrix, where for each cell we will hold, at which moment the robot visits it for the first time while moving through its route. To find these values, let’s follow all the route. Each time we move to a cell we never visited before, we must save to the corresponding matrix’ cell, how many actions are done now. Let’s prepare an array of counters, in which for each possible number of actions we will hold how many variants there were, when robot explodes after this number of actions.
Now let’s iterate through all possible cells where mine could be placed. For each cell, if it wasn’t visited by robot, add one variant of N actions, where N is the total length of the route. If it was, add one variant of that many actions as written in this cell (the moment of time when it was visited first). Look, if there is a mine in this cell, robot would explode just after first visiting it.
The array of counters is now the answer to the problem.
605A — Sorting Railway Cars. Let’s suppose we removed from the array all that elements we would move. What remains? The sequence of the numbers in a row: a, a+1, …, b. The length of this sequence must be maximal to minimize the number of elements to move. Consider the array pos, where pos[p[i]] = i. Look at it’s subsegment pos[a], pos[a+1], …, pos[b]. This sequence must be increasing and its length as mentioned above must be maximal.
So we must find the longest subsegment of pos, where pos[a], pos[a+1], …, pos[b] is increasing.
605B — Lazy Student. Let’s order edges of ascending length, in case of a tie placing earlier edges we were asked to include to MST. Let’s start adding them to the graph in this order. If we asked to include the current edge to MST, use this edge to llink 1st vertex with the least currently isolated vertex. If we asked NOT to include the current edge to MST, use this edge to link some vertices that are already linked but have no edges between them. To do this it’s convenient to have two pointer on vertices (let’s call them FROM and TO). At the beginning, FROM=2, TO=3. When we are to link two already linked vertices, we add new edge (FROM, TO) and increment FROM. If FROM becomes equal to TO, we can assume we already added all possible edges to TO, so we increment TO and set FROM to 2. This means from this moment we will use non-MST edges to connect TO with all previous vertices starting from 2. If it appears that TO looks at currently isolated vertex, we can assume there are no place for non-MST edge it the graph, so the answer is Impossible. Keep doing in the described way, we’ll be adding MST edges as (1,2), …, (1,n) and non-MST edges as (2,3), (2,4), (3,4), (2,5), (3,5), (4,5), ...
605C — Freelancer's Dreams. We can let our hero not to receive money or experience for some projects. This new opportunity does not change the answer. Consider the hero spent time T to achieve his dream. On each project he spent some part of this time (possibly zero). So the average speed of making money and experience was linear combination of speeds on all these projects, weighted by parts of time spent for each of the projects.
Let’s build the set P on the plane of points (x, y) such that we can receive x money and y experience per time unit. Place points (a[i], b[i]) on the plane. Add also two points (max(a[i]), 0) and (0, max(b[i])). All these points for sure are included to P. Find their convex hull. After that, any point inside or at the border of the convex hull would correspond to usage of some linear combination of projects.
Now we should select some point which hero should use as the average speed of receiving money and experience during all time of achieving his dream. This point should be non-strictly inside the convex hull. The dream is realized if we get to point (A,B). The problem lets us to get upper of righter, but to do so is not easier than to get to the (A,B) itself. So let’s direct a ray from (0,0) to (A,B) and find the latest moment when this ray was inside our convex hull. This point would correspond to the largest available speed of receiving resources in the direction of point (A,B). Coordinates of this point are speed of getting resources.
To find the point, we have to intersect the ray and the convex hull.
605D — Board Game. Consider n vectors starting at points (a[i], b[i]) and ending at points (c[i], d[i]). Run BFS. On each of its stages we must able to perform such an operation: get set of vectors starting inside rectangle 0 <= x <= c[i], 0 <= y <= d[i] and never consider these vectors again. It can be managed like this. Compress x-coordinates. For each x we’ll hold the list of vectors which first coordinate is x. Create a segment tree with first coordinate as index and second coordinate as value. The segment tree must be able to find index of minimum for segment and to set value at point. Now consider we have to find all the vectors with first coordinate from 0 to x and second coordinate from 0 to y. Let’s find index of minimum in the segment tree for segment [0, x]. This minimum points us to the vector (x,y), whose x — that index of minimum and y — value of minimum. Remove it from list of vectors (adding also to the queue of the BFS) and set in the segment tree to this index second coordinate of the next vector with first coordinate x. Continue this way while minimum on a segment remains less than y. So, on each step we will find list of not yet visited vectors in the bottom right rectangle, and each vector would be considered only once, after what it would be deleted from data structures.
605E — Intergalaxy Trips. The vertex is the better, the less is the expected number of moves from it to reach finish. The overall strategy is: if it is possible to move to vertex better than current, you should move to it, otherwise stay in place. Just like in Dijkstra, we will keep estimates of answer for each vertex, and fix these estimates as the final answer for all vertices one by one, starting from best vertices to the worst. On the first step we will fix vertex N (the answer for it is zero). On the second step – vertex from which it’s easiest to reach N. On the third step – vertex from which it’s easiest to finish, moving to vertices determined on first two steps. And so on. On each step we find such vertex which gives best expected number of moves if we are to move from it to vertices better than it and then we fix this expected number – it cannot change from now. For each non-fixed yet vertex we can find an estimate of expected time it takes to reach finish from it. In this estimate we take into account knowledge about vertices we know answer for. We iterate through vertices in order of non-increasing answer for them, so the answer for vertex being estimated is not better than for vertices we already iterate through. Let’s see the expression for expected time of getting to finish from vertex x, considering use of tactic “move to best of i accessible vertices we know answer for, or stay in place”:
m(x) = p(x, v) * ans(v) + (1 — p(x, v) * p(x, v) * ans(v) + (1 — p(x, v) * (1 — p(x, v) * p(x, v) * ans(v) + … + (1 — p(x, v) * (1 — p(x, v) * … * (1 — p(x, v[i-1]) * m(x) + 1
Here m(x) – estimate for vertex x, p(a,b) – the probability of existence of edge (a,b), and ans(v) – known answer for vertex v.
Note that m(x) expressed by itself, because there is a probability of staying in place.
We will keep estimating expression for each vertex in the form of m(x) = A[x] * m(x) + B[x].
For each vertex we will keep A[x] and B[x]. This would mean that with some probabilites it would be possible to move to some better vertex, and this opportunity gives contribution to expected time equal to B[x], and also with some probability we have to stay in place, and this probability is A[x] (this is just the same as coefficient before m(x) in the expression).
So, on each step we select one currently non-fixed vertex v with minimal estimate, then fix it and do relaxation from it, refreshing estimates for other vertices. When we refresh estimate for some vertex x, we change its A[x] and B[x]. A[x] is reduced by A[x] * p(x,v), because the probability of staying still consider it’s not possible to move to v. B[x] is increased by A[x] * p(x,v) * ans(v), where A[x] is the probability that it’s not possible to use some vertex better than v, A[x] * p(x,v) is the probability that it’s also possible to use vertex v, and ans(v) – known answer we just fixed for vertex v. To calculate the value of estimate for some vertex x, we can use expression m(x) = A[x] * m(x) + B[x] and express m(x) from it. Exactly m(x) is that value we should keep on the priority queue in out Dijkstra analogue, and exactly m(x) is the value to fix as the final answer for vertex x, when this vertex is announced as vertex with minimal estimate at the start of a step.