<p>You are given an undirected weighted graph of&nbsp;<code>n</code>&nbsp;nodes (0-indexed), represented by an edge list where&nbsp;<code>edges[i] = [a, b]</code>&nbsp;is an undirected edge connecting the nodes&nbsp;<code>a</code>&nbsp;and&nbsp;<code>b</code>&nbsp;with a probability of success of traversing that edge&nbsp;<code>succProb[i]</code>.</p>

<p>Given two nodes&nbsp;<code>start</code>&nbsp;and&nbsp;<code>end</code>, find the path with the maximum probability of success to go from&nbsp;<code>start</code>&nbsp;to&nbsp;<code>end</code>&nbsp;and return its success probability.</p>

<p>If there is no path from&nbsp;<code>start</code>&nbsp;to&nbsp;<code>end</code>, <strong>return&nbsp;0</strong>. Your answer will be accepted if it differs from the correct answer by at most <strong>1e-5</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/09/20/1558_ex1.png" style="width: 187px; height: 186px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.2], start = 0, end = 2
<strong>Output:</strong> 0.25000
<strong>Explanation:</strong>&nbsp;There are two paths from start to end, one having a probability of success = 0.2 and the other has 0.5 * 0.5 = 0.25.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/09/20/1558_ex2.png" style="width: 189px; height: 186px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.3], start = 0, end = 2
<strong>Output:</strong> 0.30000
</pre>

<p><strong class="example">Example 3:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/09/20/1558_ex3.png" style="width: 215px; height: 191px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[0,1]], succProb = [0.5], start = 0, end = 2
<strong>Output:</strong> 0.00000
<strong>Explanation:</strong>&nbsp;There is no path between 0 and 2.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 10^4</code></li>
	<li><code>0 &lt;= start, end &lt; n</code></li>
	<li><code>start != end</code></li>
	<li><code>0 &lt;= a, b &lt; n</code></li>
	<li><code>a != b</code></li>
	<li><code>0 &lt;= succProb.length == edges.length &lt;= 2*10^4</code></li>
	<li><code>0 &lt;= succProb[i] &lt;= 1</code></li>
	<li>There is at most one edge between every two nodes.</li>
</ul>

class Solution {
public:
    double maxProbability(int n, vector<vector<int>>& edges, vector<double>& succProb, int start, int end) {
        /*
        using Dijkstra Algorithm
        reference: https://leetcode.com/problems/path-with-maximum-probability/solutions/731767/java-python-3-2-codes-bellman-ford-and-dijkstra-s-algorithm-w-brief-explanation-and-analysis/
        */
        vector<vector<pair<int, int>>> neighbor(n); // {node index, index in edges}
        vector<double> prob(n, 0.0); // prob[i] stores the maximum probability of the path from start to i
        priority_queue<pair<double, int>> pq;
        for(int i = 0; i < edges.size(); i++){
            neighbor[edges[i][0]].push_back({edges[i][1], i});
            neighbor[edges[i][1]].push_back({edges[i][0], i});
        }
        prob[start] = 1.0;
        pq.push({prob[start], start});
        while(!pq.empty()){
            pair<double, int> top = pq.top();
            double p = top.first;
            int id = top.second;
            pq.pop();
            if(id == end)
                return p;
            for(int i = 0; i < neighbor[id].size(); i++){
                if(p * succProb[neighbor[id][i].second] > prob[neighbor[id][i].first]){
                    prob[neighbor[id][i].first] = p * succProb[neighbor[id][i].second];
                    pq.push({prob[neighbor[id][i].first], neighbor[id][i].first});
                }
            }
        }
        return 0.0;
    }
};
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