October 3, 2023<\/p>\n
Author: AI Writer<\/p>\n<\/header>\n The Kevin Bacon rule of six degrees states that any person is connected to Kevin Bacon within a maximum of six degrees. This is an intriguing concept that explains relationships in social networks and is a popular topic in various fields. In this article, we will transform this rule into an algorithmic problem and detail the solution process using C++.<\/p>\n The problem we will tackle is “Finding the person who can reach Kevin Bacon the fastest.” Through research, we will calculate the degree of connection to Kevin Bacon for each person when given the friendship relations among several people.<\/p>\n<\/section>\n The objective is to calculate each person’s Kevin Bacon number within a given network and find the person with the lowest number. To do this, we need to input the number of people and their friendship relations.<\/p>\n The input is provided in the following format:<\/p>\n The output should be the number of the person closest to Kevin Bacon. If there are multiple, output the smallest number.<\/p>\n<\/section>\n To solve this problem, we will use graph traversal techniques. We will treat people as vertices and friendships as edges, applying BFS (Breadth-First Search). We will calculate the shortest distance from each person to Kevin Bacon and find the minimum among them.<\/p>\n Now, let\u2019s implement the C++ code based on the algorithm above.<\/p>\n The above C++ code includes the following key steps:<\/p>\n The time complexity of this algorithm is O(N + M). This is because each person explores their friends through BFS, and each edge is explored only once. This is sufficiently efficient for the input range of the given problem.<\/p>\n<\/section>\n In this lecture, we explored how to solve an algorithm problem utilizing the Kevin Bacon rule of six degrees in C++. We analyzed relationships within a network using graph traversal techniques like BFS. Such types of problems frequently appear in coding tests, making it crucial to practice varied problems to enhance proficiency.<\/p>\n Finally, the problem-solving process presented here will not only aid in preparation for coding tests but also help address various issues that may arise in actual programming more systematically.<\/p>\n<\/section>\nIntroduction<\/h2>\n
Problem Definition<\/h2>\n
Problem Description<\/h3>\n
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Problem Solving Approach<\/h2>\n
Algorithm<\/h3>\n
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C++ Implementation<\/h2>\n
\n#include <iostream>\n#include <vector>\n#include <queue>\n#include <limits> \nusing namespace std;\n\nint main() {\n int N, M;\n cin >> N >> M; \/\/ Number of people N, number of friendship relations M\n vector<vector<int>> graph(N + 1); \/\/ Adjacency list\n\n for (int i = 0; i < M; i++) {\n int A, B;\n cin >> A >> B; \/\/ Friendship relation A, B\n graph[A].push_back(B);\n graph[B].push_back(A); \/\/ Undirected graph\n }\n\n int min_bacon = numeric_limits<int>::max();\n int answer = 0;\n\n for (int i = 1; i <= N; i++) {\n vector<int> distance(N + 1, -1); \/\/ Initialize distances for each person\n queue<int> q;\n distance[i] = 0; \/\/ Distance from self is 0\n q.push(i); \/\/ Add self to the queue\n\n while (!q.empty()) {\n int current = q.front();\n q.pop();\n for (int neighbor : graph[current]) {\n if (distance[neighbor] == -1) {\n distance[neighbor] = distance[current] + 1; \/\/ Calculate distance\n q.push(neighbor); \/\/ Add friend to explore\n }\n }\n }\n\n int bacon_count = 0;\n for (int d : distance) {\n if (d != -1) {\n bacon_count += d; \/\/ Sum Kevins Bacon numbers\n }\n }\n\n if (bacon_count < min_bacon) {\n min_bacon = bacon_count; \/\/ Update minimum\n answer = i; \/\/ Record person number\n }\n }\n\n cout << answer << endl; \/\/ Output result\n return 0;\n}\n <\/code><\/pre>\n<\/section>\nCode Explanation<\/h2>\n
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Complexity Analysis<\/h2>\n
Conclusion<\/h2>\n