leetcode1514 最大概率路径（Bellman-ford算法详解）

题目描述：

You are given an undirected weighted graph of n nodes (0-indexed), represented by an edge list where edgesi = a, b is an undirected edge connecting the nodes a and b with a probability of success of traversing that edge succProbi.

Given two nodes start and end, find the path with the maximum probability of success to go from start to end and return its success probability.

If there is no path from start to end, return 0. Your answer will be accepted if it differs from the correct answer by at most 1e-5.
题目链接

解题思路：

为了解决这个问题，我们需要在无向图中找到两个节点之间的路径，以最大化边概率的乘积。Bellman-Ford算法通常用于在具有负权重的图中找到最短路径，可以用来解决这个问题。我们将通过迭代更新起始点到达每个节点的最大概率来求最终的最大概率。
Bellman-Ford算法

代码实现：

package practise;

public class leetcode1514 {
    public static void main(String[] args) {
        int[][] edges = {{2,3},{1,2},{3,4},{1,3},{1,4},{0,1},{2,4},{0,4},{0,2}};
        double[] succProb = {0.06,0.26,0.49,0.25,0.2,0.64,0.23,0.21,0.77};
        System.out.println(maxProbability(5, edges, succProb, 0, 3));
    }

    public static double maxProbability(int n, int[][] edges, double[] succProb, int start_node, int end_node) {
        double[] maxProb = new double[n]; //the pro from start_node to xxx
        maxProb[start_node] = 1.0;
        for (int i = 0; i < edges.length; i++) {
            boolean updated = false;
            for (int j = 0; j < edges.length; j++) {
                int from = edges[j][0], to = edges[j][1];
                double pathProb = succProb[j];
                if (maxProb[from] * pathProb > maxProb[to]) {
                    maxProb[to] = maxProb[from] * pathProb;
                    updated = true;
                }
                if (maxProb[to] * pathProb > maxProb[from]) {
                    maxProb[from] = maxProb[to] * pathProb;
                    updated = true;
                }
            }
            if(!updated) {
                break;
            }
        }
        return maxProb[end_node];
    }
}