CSES - Course Schedule II

Authors: Benjamin Qi, Paul Chen

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This problem is equivalent to Minimal Labels from Codeforces. Treat the "label" of a vertex in "Minimal Labels" as the completion time of a course in "Course Schedule II." So it suffices to solve the CF problem (editorial) and then output the inverse permutation.

C++

#include <bits/stdc++.h>
using namespace std;


int main() {
	int n, m;
	cin >> n >> m;


	vector<int> out(n + 1);           // Number of outgoing nodes
	vector<vector<int>> radj(n + 1);  // Reverse graph

Java

import java.io.*;
import java.util.*;


public class Main {
	public static void main(String[] args) {
		Kattio io = new Kattio();
		int n = io.nextInt();
		int m = io.nextInt();
		List<List<Integer>> graph = new ArrayList<>();
		for (int i = 0; i <= n; i++) { graph.add(new LinkedList<>()); }

Python

import heapq


n, m = map(int, input().split())
out = [0] * (n + 1)  # Number of outgoing nodes
radj = [[] for _ in range(n + 1)]  # Reverse graph


for _ in range(m):
	a, b = map(int, input().split())
	radj[b].append(a)
	out[a] += 1

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