Topological Sort

To review, a directed graph consists of edges that can only be traversed in one direction. Additionally, a acyclic graph defines a graph which does not contain cycles, meaning you are unable to traverse across one or more edges and return to the node you started on. Putting these definitions together, a directed acyclic graph, sometimes abbreviated as DAG, is a graph which has edges which can only be traversed in one direction and does not contain cycles.

Topological Sort

A topological sort of a directed acyclic graph is a linear ordering of its vertices such that for every directed edge $u\to v$ from vertex $u$ to vertex $v$, $u$ comes before $v$ in the ordering.

There are two common ways to topologically sort, one involving DFS and the other involving BFS.

DFS

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

#define pb push_back

int N; // Number of nodes
vector<int> graph[100000], top_sort; // Assume that this graph is a DAG
bool visited[100000];

void dfs(int node) {

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

public class CourseSchedule {
    public static ArrayList <Integer> g[];
    public static ArrayList <Integer> topo = new ArrayList < Integer > ();
    public static int N;
    public static boolean visited[];
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

Finding a Cycle

We can modify the algorithm above to return a directed cycle in case a topological sort does not exist. To find the cycle, we add each node we visit onto the stack until we detect a node already on the stack. Once we find the cycle exists, we mark the start of the cycle as not part of the stack and revursively backtrack until we reach the start node again.

#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

bool visited[(int)10e5+5], on_stack[(int)10e5+5];
vector<int> adj[(int)10e5+5];
vector<int> cycle;
int N, M;
bool dfs(int n) {

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

public class cycle {
    public static ArrayList < Integer > [] adj;
    public static boolean visited[], on_stack[];
    public static ArrayList < Integer > cycle;
    public static int N, M;
    public static void main(String[] args) throws IOException {
        BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));

BFS

The BFS version, known as Kahn's Algorithm, makes it obvious how to extract the lexicographically minimum topological sort.

int in_degree[100000];
vector<int> edge[100000];

int N; //number of nodes

void compute() {
    queue<int> q;
    for (int i = 0; i < N; i++) {
        if (in_degree[i] == 0) {
            q.push(i);

static int in_degree[];
static ArrayList<Integer> edge[]; //adjacency list

static int N; //number of nodes

static void topological_sort() {
    Queue<Integer> q = new ArrayDeque<Integer>();
    for (int i = 0; i < N; i++) {
        if (in_degree[i] == 0) {
            q.add(i);

Dynamic Programming

One useful property of directed acyclic graphs is, as the name suggests, that no cycles exist. If we consider each node in the graph as a state, we can perform dynamic programming on the graph if we process the states in an order that guarantees for every edge $u\to v$ that $u$ is processed before $v$. Fortunately, this is the exact definition of a topological sort!

In this task, we must find the longest path in a DAG.

Problems