<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) {
        vector<vector<pair<int, float>>> graph(n);
        {
            vector<int> edge_counts(n);
            for (const auto& e : edges) {
                ++edge_counts[e.front()];
                ++edge_counts[e[1]];
            }
            auto* gp = &graph.front() - 1;
            auto* ecp = &edge_counts.front() - 1;
            for (int i = 0; i < n; ++i) {
                (++gp)->reserve(*++ecp);
            }
        }
        {
            int es = edges.size();
            double *spp = &succProb.front() - 1;
            auto *ep = &edges.front() -1;
            for (int i = 0; i < es; i++) {
                double w = *++spp;
                if (w == 0.0)
                    continue;
                const auto& e = *++ep;
                int a = e.front();
                int b = e[1];
                graph[a].emplace_back(b, w);
                graph[b].emplace_back(a, w);
            }
        }

        vector<float> dist(n, 0.0);
        dist[start] = 1.0;
        priority_queue<pair<float, int>> heap;
        heap.emplace(1.0, start);
        float du, product;
        while (!heap.empty()) {
            auto [d, u] = heap.top();
            if (d == end)
                return dist[end];
            heap.pop();
            if (d < (du = dist[u])) {
                continue;
            }
            for (auto [v, w] : graph[u]) {
                auto& dv = dist[v];
                if (dv > d || (product = du * w) < 1e-5)
                    continue;
                if (dv < product) {
                    dv = product;
                    heap.emplace(dv, v);
                }
            }
        }

        return dist[end];
    }
};

auto fast = (ios::sync_with_stdio(0), cout.tie(0), cin.tie(0), true);
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