Give this problem a try 1249 B2. Books Exchange (hard version).
Approach
Example Code
import java.io.*;
import java.util.*;
public class Solution {
static class DSU {
int[] parent;
int[] sz;
public DSU(int N) {
parent = new int[N];
sz = new int[N];
for (int i = 0; i < N; i++) {
parent[i] = i;
sz[i] = 1;
}
}
public int find(int v) {
if (parent[v] == v) return parent[v];
return parent[v] = find(parent[v]); // path compression
}
public void union(int a, int b) {
a = find(a);
b = find(b);
if (a != b) {
if (sz[a] <= sz[b]) {
int tmp = a;
a = b;
b = tmp;
}
parent[b] = a;
sz[a] += sz[b];
}
}
public int query(int v) {
return sz[v];
}
}
public static void main(String[] args) throws Exception {
// FastScanner io = new FastScanner("");
FastScanner io = new FastScanner();
int q = io.nextInt();
for (int i = 0; i < q; i++) {
int n = io.nextInt();
DSU dsu = new DSU(n);
for (int j = 0; j < n; j++) {
int p = io.nextInt()-1;
dsu.union(j, p);
}
for (int j = 0; j < n; j++) {
int p = dsu.find(j);
int a = dsu.query(p);
if (j < n-1) io.print(a + " ");
else io.println(a);
}
}
io.close();
}
/**
RESERVED NUMBERS
*/
public static int MOD = 1_000_000_007; // prime number
public static int INF = 1_000_000_007; // infinity number
/**
DATA STRUCTURES
*/
static class MultiSet {
static TreeMap<Integer, Integer> multiset;
public MultiSet() { multiset = new TreeMap<>(); }
static void add(int x) {
multiset.putIfAbsent(x, 0);
multiset.put(x, multiset.get(x)+1);
}
static void remove(int x) {
multiset.putIfAbsent(x, 0);
multiset.put(x, multiset.get(x)-1);
if (multiset.get(x) <= 0) multiset.remove(x);
}
}
/**
IO
*/
static class FastScanner extends PrintWriter {
private BufferedReader br;
private StringTokenizer st;
// standard input
public FastScanner() { this(System.in, System.out); }
public FastScanner(InputStream i, OutputStream o) {
super(o);
st = new StringTokenizer("");
br = new BufferedReader(new InputStreamReader(i));
}
// USACO-style file input
public FastScanner(String problemName) throws IOException {
super(problemName + ".out");
st = new StringTokenizer("");
br = new BufferedReader(new FileReader(problemName + ".in"));
}
// returns null if no more input
public String next() {
try {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
return st.nextToken();
} catch (Exception e) { }
return null;
}
public int nextInt() { return Integer.parseInt(next()); }
public double nextDouble() { return Double.parseDouble(next()); }
public long nextLong() { return Long.parseLong(next()); }
}
}
Complexity Analysis
- Time Complexity: $O(Q\times N \times \alpha(N))$
- Where $\alpha(N)$ is the inverse Ackermann function, which grows very slowly. $\alpha(N) \leq 4$ for $N < 10^{600}$
- Worst case scenario of DSU would be $O(logN)$
- Space Complexity: $O(N)$
Sign-off
Congratulations on making it this far! Best of luck in your future competitions!