# Topological Sort

Authors: Benjamin Qi, Nathan Chen

Contributors: Michael Cao, Andi Qu, Andrew Wang, Qi Wang, Maggie Liu, Martin Lin

Ordering the vertices of a directed acyclic graph such that each vertex is visited before its children.

To review, a **directed** graph consists of edges that can only be traversed in
one direction. Additionally, an **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

Focus Problem – try your best to solve this problem before continuing!

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.

Resources | ||||
---|---|---|---|---|

CSA | interactive, both versions |

## DFS

Resources | ||||
---|---|---|---|---|

CPH | example walkthrough | |||

CP2 | code | |||

cp-algo | code |

C++

#include <algorithm>#include <iostream>#include <vector>using std::cout;using std::endl;using std::vector;vector<int> top_sort;vector<vector<int>> graph;

Java

import java.io.*;import java.util.*;public class CourseSchedule {private static List<Integer>[] graph;private static List<Integer> topSort = new ArrayList<>();private static boolean[] visited;public static void main(String[] args) throws IOException {BufferedReader read = new BufferedReader(new InputStreamReader(System.in));

Python

import sysMAX_N = 10**5sys.setrecursionlimit(MAX_N)def dfs(node: int) -> None:for next_ in graph[node]:if not visited[next_]:visited[next_] = True

## BFS

The BFS version is known as Kahn's Algorithm.

C++

#include <algorithm>#include <iostream>#include <queue>#include <vector>using std::cout;using std::endl;using std::vector;int main() {

Java

import java.io.*;import java.util.*;public class CourseSchedule {public static void main(String[] args) throws IOException {BufferedReader read = new BufferedReader(new InputStreamReader(System.in));StringTokenizer st = new StringTokenizer(read.readLine());int n = Integer.parseInt(st.nextToken());int m = Integer.parseInt(st.nextToken());

Python

from collections import dequen, m = map(int, input().split())graph = [[] for _ in range(n)]for _ in range(m):a, b = map(int, input().split())graph[a - 1].append(b - 1)in_degree = [0 for _ in range(n)]for nodes in graph:

### Pro Tip

If the length of the array containing the end order does not equal the number of elements that need to be sorted, then there is a cycle in the graph.

### Optional

We can also use Kahn's algorithm to extract the lexicographically minimum topological sort by breaking ties lexographically.

Although the above code does not do this, one can simply replace the `queue`

with a `priority_queue`

to implement this extension.

### Finding a Cycle

Focus Problem – try your best to solve this problem before continuing!

We can modify the DFS algorithm above to return a directed cycle in the case where 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.

For example, suppose that our stack currently consists of $s_1,s_2,\ldots,s_i$ and we then visit $u=s_j$ for some $j\le i$. If that's the case, then $s_j\to s_{j+1}\to \cdots\to s_i\to s_j$ is a cycle. We can reconstruct the cycle without explicitly storing the stack b marking $u$ as not part of the stack and recursively backtracking until $u$ is reached again.

### Warning!

This code assumes that there are no self-loops, i.e. no edges from a node to itself.

C++

#include <algorithm>#include <iostream>#include <vector>using namespace std;vector<vector<int>> graph;vector<bool> visited, on_stack;vector<int> cycle;

Java

import java.io.*;import java.util.*;public class RoundTripII {private static List<Integer>[] graph;private static boolean[] visited, onStack;private static List<Integer> cycle = new ArrayList<>();public static void main(String[] args) throws IOException {BufferedReader read = new BufferedReader(new InputStreamReader(System.in));

Python

import sysMAX_N = 10**5sys.setrecursionlimit(MAX_N)def dfs(node: int) -> bool:visited[node] = on_stack[node] = Truefor next_ in graph[node]:if on_stack[next_]:

## Dynamic Programming

Resources | ||||
---|---|---|---|---|

CPH |

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!

Focus Problem – try your best to solve this problem before continuing!

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

## Solution - Longest Flight Route

Let $dp[v]$ denote the length of the longest path ending at the node $v$. Clearly

or $1$ if $v$ is node $1$. If we process the states in topological order, it is guaranteed that $dp[u]$ will already have been computed before computing $dp[v]$.

Note that the implementation of this idea below uses Kahn's algorithm for topological sorting:

C++

#include <algorithm>#include <iostream>#include <queue>#include <vector>using std::cout;using std::endl;using std::vector;int main() {

Java

import java.io.*;import java.util.*;public class LongestFlight {public static void main(String[] args) throws IOException {BufferedReader read = new BufferedReader(new InputStreamReader(System.in));StringTokenizer st = new StringTokenizer(read.readLine());int cityNum = Integer.parseInt(st.nextToken());int flightNum = Integer.parseInt(st.nextToken());

Python

from collections import dequecity_num, flight_num = list(map(int, input().split(" ")))flights = [[] for _ in range(city_num)]back_edge = [[] for _ in range(city_num)]for _ in range(flight_num):a, b = list(map(int, input().split(" ")))a, b = a - 1, b - 1flights[a].append(b)back_edge[b].append(a)

## Problems

Status | Source | Problem Name | Difficulty | Tags | |
---|---|---|---|---|---|

CSES | Easy | ## Show TagsTopoSort | |||

Kattis | Easy | ## Show TagsTopoSort | |||

Gold | Easy | ## Show TagsTopoSort | |||

CF | Easy | ## Show TagsTopoSort | |||

CF | Easy | ## Show TagsTopoSort | |||

CF | Easy | ## Show TagsTopoSort | |||

Gold | Normal | ## Show TagsBinary Search, TopoSort | |||

CF | Hard | ## Show TagsBitwise, TopoSort | |||

CSES | Hard | ## Show TagsTopoSort |

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