BCCs and 2CCs

Catalan Numbers

DP on Trees - Combining Subtrees

Querying for the minimum and number of occurrences of minimum in a range

Counting Minimums with Segment Tree

Finding nodes that must be visited along any path.

Critical

Dynamic programming on grid boundary states.

DP on Broken Profile

Additional dynamic programming techniques and optimizations like Knuth's optimization.

Additional DP Optimizations and Techniques

Visiting all edges of a graph exactly once.

Eulerian Tours

A formula for finding the number of faces in a planar graph.

Euler's Formula

Extended Euclidean Algorithm

More Complex Operations Using FFT

Introduction to Fast Fourier Transform

Flow with Lower Bounds

A simple solution to "Robotic Cow Herd" that generalizes.

Fracturing Search

Solving games that are usually two-player to find the winner.

Game Theory

Interactive and Communication Problems

Link Cut Tree

Lagrangian Relaxation

LineContainer

Matroid Intersection

Introduces maximum flow as well as flow with lower bounds.

Maximum Flow

Triangle Inequality, Johnson's Algorithm, and Min Cost Flow

Minimum Cost Flow

Minimum Cut

Erasing from non-amortized insert-only data structures.

Offline Deletion

What if data structures could time travel?

Persistent Data Structures

Prefix Sums of Number-Theoretic Functions (Part 1)

Prefix Sums of Number-Theoretic Functions (Part 2)

Subsets of nodes in directed graphs where each node in a subset can reach each other node in the subset.

Strongly Connected Components

Returning to Bellman-Ford and Floyd-Warshall.

Shortest Paths with Negative Edge Weights

Segment Tree Beats

Slope trick refers to a way to manipulate piecewise linear convex functions. Includes a simple solution to USACO Landscaping.

Slope Trick

Max Suffix Query with Insertions Only

Knuth-Morris-Pratt and Z Algorithms (and a few more related topics).

String Searching

Suffix Automata, Suffix Trees, and Palindromic Trees

String Suffix Structures

Quickly sorting suffixes of a string and its applications

Suffix Array

Treaps

Vectorization in C++

Efficiently querying for the kth minimum element in a subarray.

Wavelet Tree

XOR Basis

Using randomized algorithms to solve problems.

Randomness

Extending Range Queries to 2D (and beyond).

2D Range Queries

Binary Jumping

Examples of how bitsets give some unintended solutions on recent USACO problems.

(Optional) Bitsets

Decomposing a tree to facilitate path computations.

Centroid Decomposition

Final tips for Platinum and additional practice problems.

Additional Practice for USACO Platinum

Smallest convex polygon containing a set of points on a grid.

Convex Hull

A way to find the maximum or minimum value of several convex functions at given points.

Convex Hull Trick

Using Divide & Conquer as a DP Optimization.

Divide & Conquer - DP

Using Divide & Conquer to answer offline or online range queries on a static array.

Divide & Conquer - SRQ

Solving subset sum problems efficiently with DP.

Sum over Subsets DP

Geometry Primitives

Heavy-Light Decomposition

Repeatedly multiplying a square matrix by itself.

Matrix Exponentiation

Small-To-Large Merging

The inclusion-exclusion principle is a counting technique that generalizes the formula for computing the size of union of n finite sets.

Inclusion-Exclusion Principle

Lazy updates on segment trees and two binary indexed trees in conjunction.

Range Update Range Query

Solving 2D grid problems using 1D range queries.

Range Queries with Sweep Line

Walking on a Segment Tree, Non-Commutative Combiner Functions

More Applications of Segment Tree

Sparse Segment Trees

Splitting up data into smaller chunks to speed up processing.

Square Root Decomposition

Sweep Line

Compressing a tree to only the necessary nodes.

Virtual Tree

Decomposing Kruskal's algorithm to solve problems about minimum/maximum edge weights.

Kruskal Reconstruction Tree

DP on Trees - Solving For All Roots

Combinatorics

Final tips for Gold and additional practice problems.

Additional Practice for USACO Gold

Incorporating custom comparators into standard library containers.

(Optional) Sorted Sets with Custom Comparators

DP problems that require iterating over subsets.

Bitmask DP

Solving a DP problem on every contiguous subarray of the original array.

Range DP

DP on Trees - Introduction

The Disjoint Set Union (DSU) data structure, which allows you to add edges to a graph and test whether two vertices of the graph are connected.

Disjoint Set Union

Finding the number of integers in a range that have a property.

Digit DP

Using the information that one integer evenly divides another.

Divisibility

Quickly testing equality of substrings or sets with a small probability of failure.

Hashing

Maintaining collections of distinct elements with hashing.

(Optional) Hashmaps

Speeding up naive recursive solutions with memoization.

Introduction to DP

Finding the next element smaller or larger than a specified key in a sorted set, and using iterators with sorted sets.

More Operations on Sorted Sets

Problems that can be modeled as filling a limited-size container with items.

Knapsack DP

Finding and using the longest increasing subsequence of an array.

Longest Increasing Subsequence

Finding a subset of the edges of a connected, undirected, edge-weighted graph that connects all the vertices to each other of minimum total weight.

Minimum Spanning Trees

Problems involving dividing the search space into two.

Meet In The Middle

Modular Arithmetic

Segment Tree, Binary Indexed Tree, and Order Statistic Tree (in C++).

Point Update Range Sum

Counting the number of "special" paths on a grid, and how some string problems can be solved using grids.

Paths on Grids

Bellman-Ford, Floyd-Warshall, and Dijkstra.

Shortest Paths with Non-Negative Edge Weights

Maintaining data over consecutive subarrays.

Sliding Window

A data structure that only allows insertion and deletion at one end.

Stacks

Using ternary or binary search to find the mode of unimodal functions.

Optimizing Unimodal Functions

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

Topological Sort

Flattening a tree into an array to easily query and update subtrees.

Euler Tour Technique

Introduces how BFS can be used to find shortest paths in unweighted graphs.

Shortest Paths with Unweighted Edges

Binary searching on arbitrary monotonic functions.

Binary Search

Efficiently searching for a value in a sorted array.

Binary Search on a Sorted Array

Final tips for Silver and additional practice problems.

Additional Practice for USACO Silver

Finding connected components in a graph represented by a grid.

Flood Fill

Directed graphs in which every vertex has exactly one outgoing edge.

Introduction to Functional Graphs

Traversing a graph with depth first search and breadth first search.

Graph Traversal

Solving greedy problems by sorting the input.

Greedy Algorithms with Sorting

Six bitwise operators and the common ways they are used.

Intro to Bitwise Operators

Introducing a special type of graph: trees.

Introduction to Tree Algorithms

Complete Search with Recursion

Introduction to Graphs

Harder problems involving iterating through the entire solution space, including those that require generating subsets and permutations.

Livestock Lineup

"use strict";
const {Fragment: _Fragment, jsx: _jsx, jsxs: _jsxs} = arguments[0];
const frontmatter = undefined;
function _createMdxContent(props) {
 const _components = {
 a: "a",
 code: "code",
 h2: "h2",
 p: "p",
 pre: "pre",
 strong: "strong",
 ...props.components
 }, {Asterisk, CPPSection, HeaderLink, JavaSection, LanguageSection, MATHSPAN, PySection, Spoiler, YouTube} = _components;
 if (!Asterisk) _missingMdxReference("Asterisk", true);
 if (!CPPSection) _missingMdxReference("CPPSection", true);
 if (!HeaderLink) _missingMdxReference("HeaderLink", true);
 if (!JavaSection) _missingMdxReference("JavaSection", true);
 if (!LanguageSection) _missingMdxReference("LanguageSection", true);
 if (!MATHSPAN) _missingMdxReference("MATHSPAN", true);
 if (!PySection) _missingMdxReference("PySection", true);
 if (!Spoiler) _missingMdxReference("Spoiler", true);
 if (!YouTube) _missingMdxReference("YouTube", true);
 return _jsxs(_Fragment, {
 children: [_jsx(_components.p, {
 children: _jsx(_components.a, {
 href: "http://www.usaco.org/current/data/sol_lineup_bronze_dec19.html",
 rel: "nofollow",
 target: "_blank",
 children: "Official Analysis (C++)"
 })
 }), "\n", _jsxs(_components.h2, {
 id: "video-solution",
 children: [_jsx(_components.a, {
 "aria-hidden": "true",
 tabIndex: "-1",
 className: "anchor before",
 href: "#video-solution",
 children: _jsx(HeaderLink, {})
 }), "Video Solution"]
 }), "\n", _jsx(_components.p, {
 children: "By Melody Yu"
 }), "\n", _jsx(YouTube, {
 id: "xed_5FjVMoo"
 }), "\n", _jsx(Spoiler, {
 title: "Video Solution Code",
 children: _jsxs(LanguageSection, {
 children: [_jsx(CPPSection, {
 children: _jsx(_components.pre, {
 children: _jsx(_components.code, {
 className: "language-cpp",
 children: "#include <bits/stdc++.h>\nusing namespace std;\n\nint N;\nint main() {\n\tfreopen(\"lineup.in\", \"r\", stdin);\n\tfreopen(\"lineup.out\", \"w\", stdout);\n\tcin >> N;\n\n\tvector<pair<string, string>> restrictions;\n\tfor (int i = 0; i < N; i++) {\n\t\tstring a, t, b;\n\t\tcin >> a;\n\t\tcin >> t >> t >> t >> t;\n\t\tcin >> b;\n\t\trestrictions.push_back({a, b});\n\t}\n\n\tvector<string> cows = {\"Bessie\", \"Buttercup\", \"Belinda\", \"Beatrice\",\n\t \"Bella\", \"Blue\", \"Betsy\", \"Sue\"};\n\tsort(cows.begin(), cows.end());\n\tint count = 0;\n\n\twhile (next_permutation(cows.begin(), cows.end())) {\n\t\tbool passed = true;\n\t\tfor (auto p : restrictions) {\n\t\t\tstring cow1 = p.first;\n\t\t\tstring cow2 = p.second;\n\t\t\tauto a = find(cows.begin(), cows.end(), cow1);\n\t\t\tauto b = find(cows.begin(), cows.end(), cow2);\n\t\t\tif (abs(a - b) != 1) { passed = false; }\n\t\t}\n\t\tif (passed) { break; }\n\t}\n\n\tfor (auto cow : cows) { cout << cow << endl; }\n}\n"
 })
 })
 }), _jsx(JavaSection, {
 children: _jsx(_components.pre, {
 children: _jsx(_components.code, {
 className: "language-java",
 children: "import java.io.*;\nimport java.util.*;\n\npublic class Lineup {\n\tstatic ArrayList<String> restrictionsA;\n\tstatic ArrayList<String> restrictionsB;\n\n\tprivate static ArrayList<String> permutations(ArrayList<String> s) {\n\t\t// 1. finds the largest k, that c[k] < c[k+1]\n\t\tint first = s.size() - 2;\n\t\tfor (; first >= 0; first--) {\n\t\t\tif (s.get(first).compareTo(s.get(first + 1)) < 0) break;\n\t\t}\n\n\t\tif (first == -1) return null;\n\n\t\t// 2. find last index toSwap, that c[k] < c[toSwap]\n\t\tint toSwap = s.size() - 1;\n\t\tfor (; toSwap >= 0; toSwap--) {\n\t\t\tif (s.get(first).compareTo(s.get(toSwap)) < 0) break;\n\t\t}\n\n\t\t// 3. swap elements with indexes first and last\n\t\tCollections.swap(s, first++, toSwap);\n\n\t\t// 4. reverse sequence from k+1 to n (inclusive)\n\t\ttoSwap = s.size() - 1;\n\t\twhile (first < toSwap) Collections.swap(s, first++, toSwap--);\n\n\t\treturn s;\n\t}\n\n\tprivate static int findIndex(ArrayList<String> perm, String cow) {\n\t\tfor (int i = 0; i < perm.size(); i++) {\n\t\t\tif (cow.equals(perm.get(i))) { return i; }\n\t\t}\n\t\treturn -1;\n\t}\n\n\tprivate static boolean check(ArrayList<String> perm) {\n\t\tboolean passed = true;\n\t\tfor (int i = 0; i < restrictionsA.size(); i++) {\n\t\t\tString cow1 = restrictionsA.get(i);\n\t\t\tString cow2 = restrictionsB.get(i);\n\t\t\tint a = findIndex(perm, cow1);\n\t\t\tint b = findIndex(perm, cow2);\n\t\t\tif (Math.abs(a - b) != 1) {\n\t\t\t\tpassed = false;\n\t\t\t\tbreak;\n\t\t\t}\n\t\t}\n\n\t\tif (passed) {\n\t\t\treturn true;\n\t\t} else {\n\t\t\treturn false;\n\t\t}\n\t}\n\n\tpublic static void main(String[] args) throws IOException {\n\t\tLineup.Kattio io = new Lineup.Kattio(\"lineup\");\n\t\tint N = io.nextInt();\n\n\t\tArrayList<String> cows = new ArrayList<String>(\n\t\t Arrays.asList(\"Bessie\", \"Buttercup\", \"Belinda\", \"Beatrice\", \"Bella\", \"Blue\",\n\t\t \"Betsy\", \"Sue\"));\n\n\t\trestrictionsA = new ArrayList<String>();\n\t\trestrictionsB = new ArrayList<String>();\n\n\t\tfor (int i = 0; i < N; i++) {\n\t\t\tString a = io.next();\n\t\t\tio.next();\n\t\t\tio.next();\n\t\t\tio.next();\n\t\t\tio.next();\n\t\t\tString b = io.next();\n\t\t\trestrictionsA.add(a);\n\t\t\trestrictionsB.add(b);\n\t\t}\n\n\t\tCollections.sort(cows);\n\t\twhile (cows != null) {\n\t\t\tif (check(cows)) {\n\t\t\t\tfor (String c : cows) { io.println(c); }\n\t\t\t\tbreak;\n\t\t\t}\n\t\t\tcows = permutations(cows);\n\t\t}\n\n\t\tio.close();\n\t}\n\n\t//BeginCodeSnip{Kattio}\n\tstatic class Kattio extends PrintWriter {\n\t\tprivate BufferedReader r;\n\t\tprivate StringTokenizer st;\n\t\t// standard input\n\t\tpublic Kattio() { this(System.in, System.out); }\n\t\tpublic Kattio(InputStream i, OutputStream o) {\n\t\t\tsuper(o);\n\t\t\tr = new BufferedReader(new InputStreamReader(i));\n\t\t}\n\t\t// USACO-style file input\n\t\tpublic Kattio(String problemName) throws IOException {\n\t\t\tsuper(problemName + \".out\");\n\t\t\tr = new BufferedReader(new FileReader(problemName + \".in\"));\n\t\t}\n\t\t// returns null if no more input\n\t\tpublic String next() {\n\t\t\ttry {\n\t\t\t\twhile (st == null || !st.hasMoreTokens())\n\t\t\t\t\tst = new StringTokenizer(r.readLine());\n\t\t\t\treturn st.nextToken();\n\t\t\t} catch (Exception e) { }\n\t\t\treturn null;\n\t\t}\n\t\tpublic int nextInt() { return Integer.parseInt(next()); }\n\t\tpublic double nextDouble() { return Double.parseDouble(next()); }\n\t\tpublic long nextLong() { return Long.parseLong(next()); }\n\t}\n\t//EndCodeSnip\n}\n"
 })
 })
 })]
 })
 }), "\n", _jsxs(_components.h2, {
 id: "explanation",
 children: [_jsx(_components.a, {
 "aria-hidden": "true",
 tabIndex: "-1",
 className: "anchor before",
 href: "#explanation",
 children: _jsx(HeaderLink, {})
 }), "Explanation"]
 }), "\n", _jsxs(_components.p, {
 children: ["Since there are only ", _jsx(MATHSPAN, {
 className: "language-math math-inline",
 latex: "8",
 children: "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mn>8</mn></mrow><annotation encoding=\"application/x-tex\">8</annotation></semantics></math>8"
 }), " cows, there are ", _jsx(MATHSPAN, {
 className: "language-math math-inline",
 latex: "8! = 40320",
 children: "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mn>8</mn><mo stretchy=\"false\">!</mo><mo>=</mo><mn>40320</mn></mrow><annotation encoding=\"application/x-tex\">8! = 40320</annotation></semantics></math>8!=40320"
 }), " distinct possible orderings, which is small enough for us to try all of them and still pass the problem in time."]
 }), "\n", _jsx(_components.p, {
 children: "If we generate them in alphabetical order and find an ordering that satisfies all the given constraints, then we can stop and print out the answer right then and there."
 }), "\n", _jsxs(_components.h2, {
 id: "implementation",
 children: [_jsx(_components.a, {
 "aria-hidden": "true",
 tabIndex: "-1",
 className: "anchor before",
 href: "#implementation",
 children: _jsx(HeaderLink, {})
 }), "Implementation"]
 }), "\n", _jsxs(_components.p, {
 children: [_jsx(_components.strong, {
 children: "Time Complexity:"
 }), " ", _jsx(MATHSPAN, {
 className: "language-math math-inline",
 latex: "\\mathcal{O}(N)",
 children: "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mi mathvariant=\"script\">O</mi><mo stretchy=\"false\">(</mo><mi>N</mi><mo stretchy=\"false\">)</mo></mrow><annotation encoding=\"application/x-tex\">\\mathcal{O}(N)</annotation></semantics></math>O(N)"
 }), _jsx(Asterisk, {
 children: "Note that we don't have to include the time it takes to build permutations and iterate through them because they're constant factors."
 })]
 }), "\n", _jsxs(LanguageSection, {
 children: [_jsx(CPPSection, {
 children: _jsx(_components.pre, {
 children: _jsx(_components.code, {
 className: "language-cpp",
 children: "#include <bits/stdc++.h>\n\nusing namespace std;\n\nconst int RESTRICT_LEN = 6;\n// list of cows, in alphabetical order\nconst vector<string> COWS = {\"Beatrice\", \"Belinda\", \"Bella\", \"Bessie\",\n \"Betsy\", \"Blue\", \"Buttercup\", \"Sue\"};\n\nvector<vector<string>> orderings;\nvoid build(vector<string> ordering) {\n\t// finished building permutation\n\tif ((int)(ordering.size()) == 8) {\n\t\torderings.push_back(ordering);\n\t\treturn;\n\t}\n\n\tfor (const string &COW : COWS) {\n\t\tif (find(ordering.begin(), ordering.end(), COW) == ordering.end()) {\n\t\t\tordering.push_back(COW);\n\t\t\tbuild(ordering);\n\t\t\tordering.pop_back();\n\t\t}\n\t}\n}\n\n// returns index of a cow in an ordering\nint loc(const vector<string> &order, const string &cow) {\n\treturn find(order.begin(), order.end(), cow) - order.begin();\n}\n\nint main() {\n\tfreopen(\"lineup.in\", \"r\", stdin);\n\tfreopen(\"lineup.out\", \"w\", stdout);\n\tint n;\n\tcin >> n;\n\n\tvector<pair<string, string>> restrictions;\n\tfor (int i = 0; i < n; i++) {\n\t\tstring cow1 = \"\";\n\t\tstring cow2 = \"\";\n\t\tfor (int j = 0; j < RESTRICT_LEN; j++) {\n\t\t\tstring word;\n\t\t\tcin >> word;\n\t\t\tcow1 = cow1.empty() ? word : cow1;\n\t\t\tcow2 = word;\n\t\t}\n\t\trestrictions.emplace_back(cow1, cow2);\n\t}\n\n\t// build all possible orderings of cows\n\tbuild({});\n\tfor (vector<string> &order : orderings) {\n\t\tbool ok = true;\n\t\tfor (const pair<string, string> &rule : restrictions) {\n\t\t\tif (abs(loc(order, rule.first) - loc(order, rule.second)) > 1) {\n\t\t\t\tok = 0;\n\t\t\t\tbreak;\n\t\t\t}\n\t\t}\n\n\t\tif (ok) {\n\t\t\tfor (const string &i : order) { cout << i << '\\n'; }\n\t\t\tbreak;\n\t\t}\n\t}\n}\n"
 })
 })
 }), _jsx(JavaSection, {
 children: _jsx(_components.pre, {
 children: _jsx(_components.code, {
 className: "language-java",
 children: "import java.io.*;\nimport java.util.*;\n\nclass Pair<T, U> {\n\tT first;\n\tU second;\n\n\tpublic Pair(T first, U second) {\n\t\tthis.first = first;\n\t\tthis.second = second;\n\t}\n}\n\npublic class Lineup {\n\tprivate static final int RESTRICT_LEN = 6;\n\t// list of cows, in alphabetical order\n\tprivate static final List<String> COWS = Arrays.asList(\n\t \"Beatrice\", \"Belinda\", \"Bella\", \"Bessie\", \"Betsy\", \"Blue\", \"Buttercup\", \"Sue\");\n\n\tprivate static List<List<String>> orderings = new ArrayList<>();\n\n\tprivate static void build(List<String> ordering) {\n\t\t// finished building permutation\n\t\tif (ordering.size() == 8) {\n\t\t\torderings.add(new ArrayList<>(ordering));\n\t\t\treturn;\n\t\t}\n\n\t\tfor (String cow : COWS) {\n\t\t\tif (!ordering.contains(cow)) {\n\t\t\t\tordering.add(cow);\n\t\t\t\tbuild(ordering);\n\t\t\t\tordering.remove(ordering.size() - 1);\n\t\t\t}\n\t\t}\n\t}\n\n\t// returns index of a cow in an ordering\n\tprivate static int loc(List<String> order, String cow) {\n\t\treturn order.indexOf(cow);\n\t}\n\n\tpublic static void main(String[] args) throws IOException {\n\t\tBufferedReader br = new BufferedReader(new FileReader(\"lineup.in\"));\n\n\t\tint n = Integer.parseInt(br.readLine().trim());\n\t\tList<Pair<String, String>> restrictions = new ArrayList<>();\n\n\t\tfor (int i = 0; i < n; i++) {\n\t\t\tString[] words = br.readLine().split(\" \");\n\t\t\trestrictions.add(new Pair<>(words[0], words[words.length - 1]));\n\t\t}\n\n\t\tbr.close();\n\n\t\t// build all possible orderings of cows\n\t\tbuild(new ArrayList<>());\n\n\t\tPrintWriter pw =\n\t\t new PrintWriter(new BufferedWriter(new FileWriter(\"lineup.out\")));\n\t\tfor (List<String> order : orderings) {\n\t\t\tboolean ok = true;\n\t\t\tfor (Pair<String, String> rule : restrictions) {\n\t\t\t\tif (Math.abs(loc(order, rule.first) - loc(order, rule.second)) > 1) {\n\t\t\t\t\tok = false;\n\t\t\t\t\tbreak;\n\t\t\t\t}\n\t\t\t}\n\n\t\t\tif (ok) {\n\t\t\t\tfor (String cow : order) { pw.println(cow); }\n\t\t\t\tbreak;\n\t\t\t}\n\t\t}\n\n\t\tpw.close();\n\t}\n}\n"
 })
 })
 }), _jsx(PySection, {
 children: _jsx(_components.pre, {
 children: _jsx(_components.code, {
 className: "language-py",
 children: "from typing import List\n\n# list of cows, in alphabetical order\nCOWS = [\"Beatrice\", \"Belinda\", \"Bella\", \"Bessie\", \"Betsy\", \"Blue\", \"Buttercup\", \"Sue\"]\n\norderings = []\n\n\ndef build(ordering: List[str]):\n\t# finished building permutation\n\tif len(ordering) == 8:\n\t\torderings.append(ordering.copy())\n\t\treturn\n\n\tfor cow in COWS:\n\t\tif cow not in ordering:\n\t\t\tordering.append(cow)\n\t\t\tbuild(ordering)\n\t\t\tordering.pop()\n\n\nwith open(\"lineup.in\") as read:\n\tn = int(read.readline())\n\trestrictions = []\n\tfor _ in range(n):\n\t\tline = read.readline().split()\n\t\tcow1 = line[0]\n\t\tcow2 = line[-1]\n\t\trestrictions.append((cow1, cow2))\n\n# build all possible orderings of cows\nbuild([])\nfor order in orderings:\n\tfor rule in restrictions:\n\t\tif abs(order.index(rule[0]) - order.index(rule[1])) > 1:\n\t\t\tbreak\n\telse:\n\t\twith open(\"lineup.out\", \"w\") as written:\n\t\t\tfor cow in order:\n\t\t\t\tprint(cow, file=written)\n\t\tbreak\n"
 })
 })
 })]
 }), "\n", _jsxs(_components.h2, {
 id: "mathcalon-solution-with-graphs",
 children: [_jsx(_components.a, {
 "aria-hidden": "true",
 tabIndex: "-1",
 className: "anchor before",
 href: "#mathcalon-solution-with-graphs",
 children: _jsx(HeaderLink, {})
 }), _jsx(MATHSPAN, {
 className: "language-math math-inline",
 latex: "\\mathcal{O}(N)",
 children: "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mi mathvariant=\"script\">O</mi><mo stretchy=\"false\">(</mo><mi>N</mi><mo stretchy=\"false\">)</mo></mrow><annotation encoding=\"application/x-tex\">\\mathcal{O}(N)</annotation></semantics></math>O(N)"
 }), " Solution With Graphs"]
 }), "\n", _jsxs(_components.p, {
 children: ["This solution is covered in the ", _jsx(_components.a, {
 href: "/bronze/intro-graphs",
 children: "Introduction to Graphs"
 }), " module."]
 })]
 });
}
function MDXContent(props = {}) {
 const {wrapper: MDXLayout} = props.components || ({});
 return MDXLayout ? _jsx(MDXLayout, {
 ...props,
 children: _jsx(_createMdxContent, {
 ...props
 })
 }) : _createMdxContent(props);
}
return {
 frontmatter,
 default: MDXContent
};
function _missingMdxReference(id, component) {
 throw new Error("Expected " + (component ? "component" : "object") + " `" + id + "` to be defined: you likely forgot to import, pass, or provide it.");
}

Table of Contents

Table of Contents

Video Solution

Explanation

Implementation

$\mathcal{O}(N)$ Solution With Graphs

Join the USACO Forum!

Table of Contents

Table of Contents

Video Solution

Explanation

Implementation

O(N)\mathcal{O}(N)O(N) Solution With Graphs

Join the USACO Forum!

$\mathcal{O}(N)$ Solution With Graphs