Submission #364322

#TimeUsernameProblemLanguageResultExecution timeMemory
364322JerryLiu06Bitaro’s Party (JOI18_bitaro)Java
0 / 100
320 ms21092 KiB
import java.io.*; import java.util.*; // Solution Notes: The key constraint that we must notice is that Y_1 + Y_2 + ... + Y_Q <= N. This motivates us to find // a solution using SQRT Decomposition. We first run a Topological Sort from node 1 and calculate the maximum distance // of each node from node 1. Denote by dist[i][j] the maximum distance from node i to node j. Note that dist[i][j] = // dist[1][j] - dist[1][i], and hence can be calculated in O(1) time. The Topogical Sort can be done in O(M + N) time. // Now, we wish to find, for each node, the SQRT(N) furthest nodes from it. To do this efficiently, we sort the nodes // by their distance from node 1 and run through the first SQRT(N) nodes at each pass. The sort is done in O(N log(N)) // and the processing takes O(N SQRT(N)) time. Consider the case when Y_i >= SQRT(N). Note that there are at most SQRT(N) // of these queries, and for each one we can naively loop through the nodes in O(N) time. If Y_i <= SQRT(N), then there // must exist at least one of the furthest SQRT(N) nodes from the node in query that is not busy, so we can loop through // these furthest nodes and calculate the maximum in O(SQRT(N)). Thus, our final time complexity is O((Q + N) SQRT(N)). @SuppressWarnings("unchecked") class bitaro { static int N, M, Q, num = 0; static int[] dist = new int[100010]; static int[] comp = new int[100010]; static ArrayList<Node>[] nDist = new ArrayList[100010]; static ArrayList<Node>[] far = new ArrayList[100010]; static ArrayList<Integer>[] adj = new ArrayList[100010]; static int[] inDeg = new int[100010]; static final int INF = Integer.MAX_VALUE / 3; static final int BLOCK = 315; public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); StringTokenizer st = new StringTokenizer(br.readLine()); N = Integer.parseInt(st.nextToken()); M = Integer.parseInt(st.nextToken()); Q = Integer.parseInt(st.nextToken()); for (int i = 0; i < N; i++) { adj[i] = new ArrayList<>(); nDist[i] = new ArrayList<>(); far[i] = new ArrayList<>(); } for (int i = 0; i < M; i++) { st = new StringTokenizer(br.readLine()); int A = Integer.parseInt(st.nextToken()) - 1; int B = Integer.parseInt(st.nextToken()) - 1; adj[A].add(B); inDeg[B]++; } genDist(); for (int i = 0; i < num; i++) Collections.sort(nDist[i]); for (int i = 0; i < num; i++) for (int j = 0; j < nDist[i].size(); j++) { for (int k = 0; k <= j && k < BLOCK; k++) { far[nDist[i].get(j).node].add(new Node(nDist[i].get(k).node, nDist[i].get(k).dist)); } } for (int i = 0; i < Q; i++) { st = new StringTokenizer(br.readLine()); int T = Integer.parseInt(st.nextToken()) - 1; int Y = Integer.parseInt(st.nextToken()); int ans = -1; boolean[] mark = new boolean[N]; for (int j = 0; j < Y; j++) mark[Integer.parseInt(st.nextToken()) - 1] = true; if (Y >= BLOCK) for (int j = 0; j < N; j++) { if (comp[T] != comp[j] || dist[T] < dist[j] || mark[j]) continue; ans = Math.max(ans, dist[T] - dist[j]); } else for (Node n : far[T]) { if (!mark[n.node]) { ans = dist[T] - dist[n.node]; break; } } System.out.println(ans); } } static void genDist() { Arrays.fill(dist, -INF); Queue<Integer> q = new LinkedList<Integer>(); for (int i = 0; i < N; i++) if (inDeg[i] == 0) { q.add(i); dist[i] = 0; comp[i] = num++; } while (!q.isEmpty()) { int cur = q.poll(); for (int nxt : adj[cur]) { inDeg[nxt]--; dist[nxt] = Math.max(dist[nxt], dist[cur] + 1); if (inDeg[nxt] == 0) { q.add(nxt); comp[nxt] = comp[cur]; } } } for (int i = 0; i < N; i++) nDist[comp[i]].add(new Node(i, dist[i])); } static class Node implements Comparable<Node> { public int node, dist; public Node(int a, int b) { node = a; dist = b; } public int compareTo(Node n) { return dist - n.dist; } } }
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