<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, std::vector<std::vector<int>>& edges, 
        std::vector<double>& succProb, int start_node, int end_node) 
    {
        std::vector<std::vector<int>> paths;
        paths.resize(n);
        for (int i = 0; i < edges.size(); ++i) {
            std::vector<int>& edge = edges[i];
            paths[edge[0]].push_back(i);
            paths[edge[1]].push_back(i);
        }

        using pair     = std::pair<int, double>;
        using function = std::function<bool(pair const&, pair const&)>;
        function compare = [](pair const& left, pair const& right) {
            return left.second < right.second;
        };

        std::vector<double> past(n);
        std::unordered_set<int> hs = { start_node };
        past[start_node] = 1;
        std::priority_queue<pair, std::vector<pair>, function> queue(compare);
        pair current = { start_node, 1 };
        while (current.first != end_node) {
            std::vector<int>& nodes = paths[current.first];
            for (auto it = nodes.begin(); it != nodes.end(); ++it) {
                pair node;
                int idx = *it;
                if (succProb[idx] < 1e-5)
                    continue;
                node.first  = edges[idx][0] == current.first ? edges[idx][1] : edges[idx][0];
                node.second = current.second * /*+ std::log10(*/succProb[idx]/*)*/;
                if (hs.find(node.first) != hs.end())
                    continue;
                if (past[node.first] < node.second) {
                    past[node.first] = node.second;
                    queue.push(node);
                }
            }
            if (queue.empty())
                break;
            current = queue.top();
            hs.insert(current.first);
            queue.pop();
        }
        double ans = 0;
        if (current.first == end_node)
            ans = /*std::pow(10, */current.second/*)*/;
        return ans;
    }
};
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