Adding Solutions

Basic Debugging

Choosing a Language

Contest Strategy

Contests

Contributing

C++ With the Command Line

Data Types

How to Debug

Debugging C++

Expected Knowledge

Fast Input & Output

(Optional) C++ - Writing Generic Code

Input & Output

Introduction to Competitive Programming

(Optional) C++ - Lambda Expressions

Resources: Learning to Code

Introducing Modules

Olympiads

How to Practice

Resources: Competitive Programming

Resources: USA-Specific

Running Code Locally

Running Code Online

USACO Camp

USACO FAQs

USACO Monthlies

Using This Guide

Working With MDX

Ad Hoc Problems

Casework

Complete Search with Recursion

Additional Practice for USACO Bronze

Basic Complete Search

Introduction to Data Structures

Introduction to Graphs

Introduction to Greedy Algorithms

Introduction to Sets & Maps

Introduction to Sorting

Rectangle Geometry

Simulation

Time Complexity

Binary Search

Binary Search on a Sorted Array

Additional Practice for USACO Silver

Flood Fill

Introduction to Functional Graphs

Graph Traversal

Greedy Algorithms with Sorting

Intro to Bitwise Operators

Introduction to Tree Algorithms

More on Prefix Sums

Introduction to Prefix Sums

Priority Queues

Custom Comparators and Coordinate Compression

Two Pointers

DP on Trees - Solving For All Roots

Combinatorics

Additional Practice for USACO Gold

(Optional) Sorted Sets with Custom Comparators

Bitmask DP

Range DP

DP on Trees - Introduction

Disjoint Set Union

Digit DP

Divisibility

Hashing

(Optional) Hashmaps

Introduction to DP

More Operations on Sorted Sets

Knapsack DP

(Optional) LCA with Euler Tour

Longest Increasing Subsequence

Minimum Spanning Trees

(Optional) Meet In The Middle

Modular Arithmetic

Point Update Range Sum

Paths on Grids

Shortest Paths with Non-Negative Edge Weights

Sliding Window

Stacks

Optimizing Unimodal Functions

Topological Sort

(Optional) Euler Tour Technique

Shortest Paths with Unweighted Edges

2D Range Queries

Binary Jumping

(Optional) Bitsets

Centroid Decomposition

Additional Practice for USACO Platinum

Convex Hull

Convex Hull Trick

Divide & Conquer - DP

Divide & Conquer - SRQ

Sum over Subsets DP

Geometry Primitives

Heavy-Light Decomposition

Matrix Exponentiation

Small-To-Large Merging

Inclusion-Exclusion Principle

组合数学

Cowpatibility

"use strict";
const {Fragment: _Fragment, jsx: _jsx, jsxs: _jsxs} = arguments[0];
const frontmatter = undefined;
function _createMdxContent(props) {
 const _components = {
 a: "a",
 code: "code",
 em: "em",
 h2: "h2",
 p: "p",
 pre: "pre",
 strong: "strong",
 ...props.components
 }, {CPPSection, HeaderLink, LanguageSection, MATHSPAN, PySection, Spoiler} = _components;
 if (!CPPSection) _missingMdxReference("CPPSection", true);
 if (!HeaderLink) _missingMdxReference("HeaderLink", true);
 if (!LanguageSection) _missingMdxReference("LanguageSection", true);
 if (!MATHSPAN) _missingMdxReference("MATHSPAN", true);
 if (!PySection) _missingMdxReference("PySection", true);
 if (!Spoiler) _missingMdxReference("Spoiler", true);
 return _jsxs(_Fragment, {
 children: [_jsx(_components.p, {
 children: _jsx(_components.a, {
 href: "http://www.usaco.org/current/data/sol_cowpatibility_gold_dec18.html",
 rel: "nofollow",
 target: "_blank",
 children: "Official Analysis (C++)"
 })
 }), "\n", _jsxs(Spoiler, {
 title: "Hint",
 children: [_jsxs(_components.p, {
 children: ["Instead of trying to find the number of incompatible pairs, try to find the number of ", _jsx(_components.em, {
 children: "compatible"
 }), " pairs instead!"]
 }), _jsxs(_components.p, {
 children: ["This is known as ", _jsx(_components.a, {
 href: "https://artofproblemsolving.com/wiki/index.php/Complementary_counting",
 rel: "nofollow",
 target: "_blank",
 children: "complementary counting"
 }), "."]
 })]
 }), "\n", _jsxs(Spoiler, {
 title: "Solution",
 children: [_jsxs(_components.h2, {
 id: "explanation",
 children: [_jsx(_components.a, {
 "aria-hidden": "true",
 tabIndex: "-1",
 className: "anchor before",
 href: "#explanation",
 children: _jsx(HeaderLink, {})
 }), "Explanation"]
 }), _jsxs(_components.p, {
 children: ["Since there are a total of ", _jsx(MATHSPAN, {
 className: "language-math math-inline",
 latex: "\\frac{N(N-1)}{2}",
 children: "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><semantics><mrow><mfrac><mrow><mi>N</mi><mo stretchy=\"false\">(</mo><mi>N</mi><mo>−</mo><mn>1</mn><mo stretchy=\"false\">)</mo></mrow><mn>2</mn></mfrac></mrow><annotation encoding=\"application/x-tex\">\\frac{N(N-1)}{2}</annotation></semantics></math>2N(N−1)​"
 }), " pairs of cows, if we can count the number of compatible pairs, we can subtract it from the total to get our answer!"]
 }), _jsxs(_components.p, {
 children: ["We can use ", _jsx(_components.a, {
 href: "https://artofproblemsolving.com/wiki/index.php/Principle_of_Inclusion-Exclusion",
 rel: "nofollow",
 target: "_blank",
 children: "PIE (Principle of Inclusion/Exclusion)"
 }), " to calculate the number of compatible pairs! We'll have to add all pairs of cows that share one flavor in common, subtract cows that share two flavors in common,\nand so on..."]
 }), _jsx(_components.p, {
 children: "To understand why this works, imagine we only have three flavors of icecreams and that some cows share more than one flavor. If we were to only\ncount the pairs of cows with one flavor, the pairs of cows that share two flavors would be overcounted, as shown in the animation below:"
 }), _jsx("img", {
 src: "/usaco-862-anim/PIE.gif"
 }), _jsx(_components.p, {
 children: "To avoid considering rearranged sets of the same flavors, we'll sort the favorite flavors of each cow beforehand."
 }), _jsxs(_components.h2, {
 id: "implementation",
 children: [_jsx(_components.a, {
 "aria-hidden": "true",
 tabIndex: "-1",
 className: "anchor before",
 href: "#implementation",
 children: _jsx(HeaderLink, {})
 }), "Implementation"]
 }), _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)"
 })]
 }), _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\nconst int FLAVORS = 5;\n\nint main() {\n\tfreopen(\"cowpatibility.in\", \"r\", stdin);\n\tfreopen(\"cowpatibility.out\", \"w\", stdout);\n\n\tlong long n;\n\tcin >> n;\n\n\tvector<array<int, FLAVORS>> cows(n);\n\n\tfor (int i = 0; i < n; i++) {\n\t\tfor (int j = 0; j < FLAVORS; j++) { cin >> cows[i][j]; }\n\n\t\t// sort flavors to avoid double counting\n\t\tsort(cows[i].begin(), cows[i].end());\n\t}\n\n\t// common[i][j] = the number of cows that share i flavors (j) in common\n\tvector<map<array<int, FLAVORS>, int>> common(FLAVORS);\n\n\tfor (int i = 0; i < n; i++) {\n\t\tcommon[4][cows[i]]++;\n\n\t\t// iterate over all subsets\n\t\tfor (int a = 0; a < FLAVORS; a++) {\n\t\t\tcommon[0][{cows[i][a]}]++;\n\t\t\tfor (int b = a + 1; b < FLAVORS; b++) {\n\t\t\t\tcommon[1][{cows[i][a], cows[i][b]}]++;\n\t\t\t\tfor (int c = b + 1; c < FLAVORS; c++) {\n\t\t\t\t\tcommon[2][{cows[i][a], cows[i][b], cows[i][c]}]++;\n\t\t\t\t\tfor (int d = c + 1; d < FLAVORS; d++) {\n\t\t\t\t\t\tcommon[3][{cows[i][a], cows[i][b], cows[i][c], cows[i][d]}]++;\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t}\n\n\tlong long compatible = 0;\n\t// use PIE to count the number of compatible pairs\n\t// (see explanation above)\n\tfor (int i = 0; i < FLAVORS; i++) {\n\t\tfor (auto &[k, v] : common[i]) {\n\t\t\tif (i % 2 == 0) {\n\t\t\t\tcompatible += (long long)v * (v - 1) / 2;\n\t\t\t} else {\n\t\t\t\tcompatible -= (long long)v * (v - 1) / 2;\n\t\t\t}\n\t\t}\n\t}\n\n\tcout << (n * (n - 1) / 2) - compatible << endl;\n}\n"
 })
 })
 }), _jsx(PySection, {
 children: _jsx(_components.pre, {
 children: _jsx(_components.code, {
 className: "language-py",
 children: "from itertools import combinations\nfrom collections import defaultdict\n\nFLAVORS = 5\n\nwith open(\"cowpatibility.in\", \"r\") as read:\n\tn = int(read.readline().strip())\n\tcows = []\n\n\tfor _ in range(n):\n\t\tflavors = list(map(int, read.readline().strip().split()))\n\t\tflavors.sort() # Sort to avoid double counting\n\t\tcows.append(flavors)\n\n# common[i][j] = the number of cows that share i flavors (j) in common\ncommon = [defaultdict(int) for _ in range(FLAVORS)]\n\nfor flavors in cows:\n\tcommon[4][tuple(flavors)] += 1\n\n\t# iterate over all subsets\n\tfor size in range(1, FLAVORS):\n\t\tfor subset in combinations(flavors, size):\n\t\t\tcommon[size - 1][subset] += 1\n\ncompatible = 0\n# use PIE to count the number of compatible pairs\n# (see explanation above)\nfor i in range(FLAVORS):\n\tfor subset, count in common[i].items():\n\t\tif i % 2 == 0:\n\t\t\tcompatible += count * (count - 1) // 2\n\t\telse:\n\t\t\tcompatible -= count * (count - 1) // 2\n\nres = n * (n - 1) // 2 - compatible\nprint(res, file=open(\"cowpatibility.out\", \"w\"))\n"
 })
 })
 })]
 })]
 })]
 });
}
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.");
}