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The Aftercow

The Aftercow. Replacing math with cow since 1993. PotW Solution (credits to Jimmy). class Edge implements Comparable<Edge> { int toNode , dist; public Edge(int a, int b ) { toNode = a; dist = b; } public int compareTo (Edge e ) { return dist - e.dist ; } }

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The Aftercow

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  1. The Aftercow Replacing math with cow since 1993

  2. PotW Solution (credits to Jimmy) class Edge implements Comparable<Edge> {int toNode, dist;public Edge(int a, int b) {toNode= a;dist = b;}public int compareTo(Edge e) {return dist - e.dist;}} Used for both storing the graph and locating closest points within the PriorityQueue in Dijkstra's

  3. PotW Solution (cont.) for (int i = 0; i < e; i++) {int node1 = input.nextInt() - 1;int node2 = input.nextInt() - 1;int dist = input.nextInt();graphThere[node1].add(new Edge(node2, dist));graphBack[node2].add(new Edge(node1, dist));}int[] distThere = dijkstra(graphThere, v, e);int[] distBack = dijkstra(graphBack, v, e); int count = 0;for (int i = 0; i < v; i++) {if (distThere[i] + distBack[i] <= money) {count++;}}System.out.println(count);

  4. PotW Solution (cont.) int[] dijkstra(ArrayList<Edge>[] graph, int v, int e) {... while (!pq.isEmpty()) { Edge e1 = pq.poll(); if (visited[e1.toNode]) continue; visited[e1.toNode] = true; for (int i = 0; i < graph[e1.toNode].size(); i++) { Edge e2 = graph[e1.toNode].get(i); int nextDist = dist[e1.toNode] + e2.dist; if (dist[e2.toNode] > nextDist) { pq.offer(new Edge(e2.toNode, nextDist)); dist[e2.toNode] = nextDist; } } } return dist; }

  5. January Silver #1: Delivery Route • N (1 <= N <= 100) farms located at different (integer) positions in the x-y plane • Farmer John wants to visit each farm in sequential order and then return to farm 1 (1, 2, 3, …, N, 1) • Takes one minute to make a step either north, south, east, or west • You can only cross each farm once • Determine the minimum amount of time it’ll take to complete entire delivery route (or -1 if no feasible delivery route is possible)

  6. Delivery Route • “Minimum time” suggests Dijkstra’s algorithm • Constructing the graph is the hard part • Create 5 nodes for each farm, one for the farm itself and the 4 adjacent coordinates • Add an edge between every pair of nodes such that it’s possible to go from (x1, y1) to (x2, y2) using a right-angle path w/o intersecting other farms • “Right-angle path” = path that changes direction at most once, moves only up/down/left/right • Runtime of O(N3) works • Use this graph w/ Dijkstra’s algorithm • Shortest path between farms is achievable using only these nodes and right-angle paths

  7. January Gold #3: Bovine Alliance • The cows in each of N farms (1 <= N <= 100,000) were instructed to build a trail to exactly one other farm (N trails total) • However, only M of these trails have been built • Farms are arguing over which farms have built trails • Calculate the number of ways the M trails could have been built, modulo 1,000,000,007.

  8. Bovine Alliance • Imagine it as a set of vertices and edges, with edges pointing to the vertex that built the edge • Count number of assignments such that each edge points to some vertex, and each vertex has at most one edge pointing to it • Solve the problem for each connected component, multiply the number of possiblities for each together • If component has n vertices, there are two different cases for the number of edges: • n-1: component is a tree, number of assignments is n • You can leave any of the n cows, and there is one solution for each such possibility • n: component is a cycle, number of assignments is 2 • The cycle can point in two directions • Use Depth-First Search to find and categorize each connected component • Count node and edges as you DFS

  9. December Gold/Silver #1: Cow Photography • Farmer John wants to take a photograph of up to 20,000 cows standing in a specific order • Every time he’s about to take a picture a group of zero or more cows moves to a set of new positions • This happens for 5 photographs • Each cow actively moves in at most one photograph • Given the contents of each photograph, reconstruct the original intended order of the cows • This sounds familiar…

  10. Cow Photography • The solution basically involves sorting the cows • Consider two cows from the set, A and B • Neither A nor B moved in at least 3 of the 5 photographs • Compare the number of times A comes before B • 3 or more: A is before B in original ordering • Else: B is before A • Use this as comparator and sort the list

  11. PotW • No Problem of the Week this week • Practice on your own! :D • Submissions for the tiebreaker “Cow Trails” problem over the December break have been graded and ranked • Will be announced next meeting, along with prizes for the top scorers of first semester!

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