求Pareto最优化和蚁群算法相结合的算法的源码,即Pareto蚁群算法 Java的
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- turtleSteps 2024-04-05 10:59关注
以下Java示例模拟了一个简单的问题,其中每个城市之间的距离由 graph 矩阵表示。在 main 方法中,创建了一个 ParetoAntColony 对象并运行了 optimize 方法来执行算法。最后,输出找到的所有解。
import java.util.ArrayList; import java.util.List; import java.util.Random; public class ParetoAntColony { private static final int ANT_COUNT = 10; private static final int MAX_ITERATIONS = 100; private static final double ALPHA = 1.0; // 控制信息素重要程度 private static final double BETA = 2.0; // 控制启发函数重要程度 private static final double RHO = 0.5; // 信息素挥发速度 private static final double Q = 500; // 信息素增加强度系数 private static final double INITIAL_PHEROMONE = 1.0; // 初始信息素浓度 private static final int DIMENSION = 2; // 问题维度 private int[][] graph; private double[][] pheromones; private double[][] heuristicValues; private List<int[]> solutions; public ParetoAntColony(int[][] graph) { this.graph = graph; this.pheromones = new double[graph.length][graph.length]; this.heuristicValues = new double[graph.length][graph.length]; this.solutions = new ArrayList<>(); initializePheromones(); initializeHeuristicValues(); } private void initializePheromones() { for (int i = 0; i < pheromones.length; i++) { for (int j = 0; j < pheromones[i].length; j++) { pheromones[i][j] = INITIAL_PHEROMONE; } } } private void initializeHeuristicValues() { for (int i = 0; i < graph.length; i++) { for (int j = 0; j < graph[i].length; j++) { heuristicValues[i][j] = 1.0 / (graph[i][j] + 0.0001); } } } public void optimize() { for (int iteration = 0; iteration < MAX_ITERATIONS; iteration++) { for (int ant = 0; ant < ANT_COUNT; ant++) { int[] solution = constructSolution(); solutions.add(solution); updatePheromones(solution); } evaporatePheromones(); } } private int[] constructSolution() { int[] solution = new int[DIMENSION]; boolean[] visited = new boolean[graph.length]; Random random = new Random(); int current = random.nextInt(graph.length); solution[0] = current; visited[current] = true; for (int i = 1; i < DIMENSION; i++) { int next = selectNext(current, visited); solution[i] = next; visited[next] = true; current = next; } return solution; } private int selectNext(int current, boolean[] visited) { double[] probabilities = new double[graph.length]; double total = 0; for (int i = 0; i < graph.length; i++) { if (!visited[i]) { probabilities[i] = Math.pow(pheromones[current][i], ALPHA) * Math.pow(heuristicValues[current][i], BETA); total += probabilities[i]; } } double rand = Math.random() * total; double sum = 0; for (int i = 0; i < probabilities.length; i++) { if (!visited[i]) { sum += probabilities[i]; if (rand <= sum) { return i; } } } return -1; // Should not reach here } private void updatePheromones(int[] solution) { for (int i = 0; i < solution.length - 1; i++) { pheromones[solution[i]][solution[i + 1]] += Q; } pheromones[solution[solution.length - 1]][solution[0]] += Q; // Close the loop } private void evaporatePheromones() { for (int i = 0; i < pheromones.length; i++) { for (int j = 0; j < pheromones[i].length; j++) { pheromones[i][j] *= (1 - RHO); } } } public List<int[]> getSolutions() { return solutions; } public static void main(String[] args) { // Example usage int[][] graph = { {0, 2, 3}, {2, 0, 4}, {3, 4, 0} }; ParetoAntColony pac = new ParetoAntColony(graph); pac.optimize(); List<int[]> solutions = pac.getSolutions(); for (int[] solution : solutions) { for (int city : solution) { System.out.print(city + " "); } System.out.println(); } } }
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