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{"id":34274,"date":"2024-11-01T09:26:16","date_gmt":"2024-11-01T09:26:16","guid":{"rendered":"http:\/\/atmokpo.com\/w\/?p=34274"},"modified":"2024-11-01T10:57:54","modified_gmt":"2024-11-01T10:57:54","slug":"c-coding-test-course-union-find-2","status":"publish","type":"post","link":"https:\/\/atmokpo.com\/w\/34274\/","title":{"rendered":"C++ Coding Test Course, Union Find"},"content":{"rendered":"

<\/p>\n

The Union-Find data structure is very useful for solving algorithmic problems. In this article, we will explain the concept of Union-Find and detail the problem-solving process using C++.<\/p>\n

What is Union-Find?<\/h2>\n

The Union-Find algorithm is a data structure that efficiently manages a collection of data. This data structure is also known as Disjoint Set and primarily supports the following two operations:<\/p>\n

Find:<\/strong> This operation finds which set a particular element belongs to. It returns the representative element (root node) of the set.<\/li>\n

Union:<\/strong> This operation merges two sets. It connects the root nodes of the two sets to form one set.<\/li>\n<\/ul>\n
It is mainly used to find cycles in graph theory or to check connected components. There are two optimization techniques to effectively utilize Union-Find:<\/p>\n
1. Path Compression<\/h3>\n
When performing the Find operation, it updates the parent of all nodes along the path to the root node, allowing for faster retrieval of the root node in subsequent operations.<\/p>\n
2. Rank-based Union<\/h3>\n
When performing the Union operation, it compares the sizes of two sets and connects the smaller set as a child of the larger set. This helps to reduce the height of the tree.<\/p>\n
Problem Description<\/h2>\n
Now, let’s look at a problem that can be solved using the Union-Find algorithm. We will solve the problem below.<\/p>\n
Problem: Friend Network<\/h3>\n
There is a friend network in which several friends are connected to each other. Each friend is identified by a number from 1 to N. Two friends are considered friends if they are directly or indirectly connected. Given a pair of friends (a, b), check whether a and b belong to the same friend group in the friend network.<\/p>\n
Input Format<\/h4>\n
\n
The first line contains the number of friends N and the number of friend relationships M. (1 \u2264 N \u2264 100,000, 1 \u2264 M \u2264 100,000)<\/li>\n
From the second line onward, M pairs of friend relationships a, b are given.<\/li>\n
The last line contains the query Q, where each query provides a pair of friends (x, y).<\/li>\n<\/ul>\n
Output Format<\/h4>\n
For each query, print ‘YES’ if x and y belong to the same friend group, otherwise print ‘NO’.<\/p>\n
Algorithm Approach<\/h2>\n
\n
Initialize and connect the friend relationships using Union-Find. Use Union(a, b)<\/code> to connect a and b in the same set.<\/li>\n
For each query, use Find(x)<\/code> and Find(y)<\/code> to check the root nodes of the two friends. If they have the same root node, print ‘YES’; otherwise, print ‘NO’.<\/li>\n<\/ol>\nC++ Code Implementation<\/h2>\nBelow is the C++ code that implements the Union-Find algorithm:<\/p>\n \n#include \n#include \nusing namespace std;\n\nclass UnionFind {\npublic:\n UnionFind(int n) {\n parent.resize(n);\n rank.resize(n, 0);\n for (int i = 0; i < n; ++i) {\n parent[i] = i;\n }\n }\n\n int find(int x) {\n if (parent[x] != x) {\n parent[x] = find(parent[x]); \/\/ Path compression\n }\n return parent[x];\n }\n\n void unionSets(int x, int y) {\n int rootX = find(x);\n int rootY = find(y);\n\n if (rootX != rootY) {\n \/\/ Rank-based union\n if (rank[rootX] > rank[rootY]) {\n parent[rootY] = rootX;\n } else if (rank[rootX] < rank[rootY]) {\n parent[rootX] = rootY;\n } else {\n parent[rootY] = rootX;\n rank[rootX]++;\n }\n }\n }\n\nprivate:\n vector parent;\n vector rank;\n};\n\nint main() {\n int N, M;\n cin >> N >> M;\n UnionFind uf(N + 1);\n\n for (int i = 0; i < M; i++) {\n int a, b;\n cin >> a >> b;\n uf.unionSets(a, b);\n }\n\n int Q;\n cin >> Q;\n for (int i = 0; i < Q; i++) {\n int x, y;\n cin >> x >> y;\n if (uf.find(x) == uf.find(y)) {\n cout << \"YES\" << endl;\n } else {\n cout << \"NO\" << endl;\n }\n }\n\n return 0;\n}\n<\/code><\/pre>\nCode Explanation<\/h2>\nIn the above C++ code, we implemented the following components:<\/p>\n \nUnionFind class:<\/strong> Contains the main logic of the Union-Find algorithm. The find<\/code> function finds the root node, and the unionSets<\/code> function merges two sets.<\/li>\n Main function:<\/strong> Takes the friend relationships as input, stores them in the Union-Find structure, and outputs the results for the given queries.<\/li>\n<\/ul>\nTime Complexity<\/h2>\nThe time complexity of the Union-Find algorithm is as follows:<\/p>\n \nFind operation: Almost constant time (\u03b1(N), proportional to the inverse Ackermann function)<\/li>\n Union operation: Almost constant time<\/li>\n<\/ul>\nThus, the overall time complexity of the algorithm is O(M * \u03b1(N)), where M is the number of friend relationships.<\/p>\n Conclusion<\/h2>\nIn this course, we have examined the concept of the Union-Find data structure and the problem-solving process using C++ in detail. Union-Find is a powerful tool that can be effectively utilized in various problems. I hope it will be of great help in your future coding tests.<\/p>\n <\/body><\/p>\n","protected":false},"excerpt":{"rendered":" The Union-Find data structure is very useful for solving algorithmic problems. In this article, we will explain the concept of Union-Find and detail the problem-solving process using C++. What is Union-Find? The Union-Find algorithm is a data structure that efficiently manages a collection of data. This data structure is also known as Disjoint Set and … \ub354 \ubcf4\uae30 “C++ Coding Test Course, Union Find”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[111],"tags":[],"class_list":["post-34274","post","type-post","status-publish","format-standard","hentry","category-c-coding-test-tutorials-2"],"yoast_head":"\nC++ Coding Test Course, Union Find - \ub77c\uc774\ube0c\uc2a4\ub9c8\ud2b8<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/atmokpo.com\/w\/34274\/\" \/>\n<meta property=\"og:locale\" content=\"ko_KR\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"C++ Coding Test Course, Union Find - \ub77c\uc774\ube0c\uc2a4\ub9c8\ud2b8\" \/>\n<meta property=\"og:description\" content=\"The Union-Find data structure is very useful for solving algorithmic problems. 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1. Path Compression<\/h3>\nWhen performing the Find operation, it updates the parent of all nodes along the path to the root node, allowing for faster retrieval of the root node in subsequent operations.<\/p>\n

2. Rank-based Union<\/h3>\nWhen performing the Union operation, it compares the sizes of two sets and connects the smaller set as a child of the larger set. This helps to reduce the height of the tree.<\/p>\n

Problem Description<\/h2>\nNow, let’s look at a problem that can be solved using the Union-Find algorithm. We will solve the problem below.<\/p>\n

Output Format<\/h4>\nFor each query, print ‘YES’ if x and y belong to the same friend group, otherwise print ‘NO’.<\/p>\n

1. Path Compression<\/h3>\n
When performing the Find operation, it updates the parent of all nodes along the path to the root node, allowing for faster retrieval of the root node in subsequent operations.<\/p>\n

2. Rank-based Union<\/h3>\n
When performing the Union operation, it compares the sizes of two sets and connects the smaller set as a child of the larger set. This helps to reduce the height of the tree.<\/p>\n

Problem Description<\/h2>\n
Now, let’s look at a problem that can be solved using the Union-Find algorithm. We will solve the problem below.<\/p>\n

Output Format<\/h4>\n
For each query, print ‘YES’ if x and y belong to the same friend group, otherwise print ‘NO’.<\/p>\n