<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_node, int end_node) {
        
        // convert all prob to log
        // convert all these log value to negative
        // apply dijkstra classical
        // find the shortest path, again negate it and de-log it.
        
        vector<vector<pair<int, double>>> graph(n);

        for(int i=0; i<edges.size(); i++) {
            double val = log2(succProb[i]);
            graph[edges[i][0]].push_back({edges[i][1], -val});
            graph[edges[i][1]].push_back({edges[i][0], -val});

        }

        // dijkstra
        priority_queue<pair<double, pair<int,int>>, vector<pair<double, pair<int,int>>>, greater<pair<double, pair<int,int>>>> pq;
        pq.push({0, {start_node, -1}});

        vector<double> dist(n, 1e8);
        vector<int> parent(n, -1);
        dist[start_node] = 0;

        while(!pq.empty()) {
            pair<double, pair<int,int>> pr = pq.top();
            pq.pop();

            int node = pr.second.first;
            int par = pr.second.second;

            parent[node] = par;

            if(node == end_node) { // early stopping
                break;
            }

            for(pair<int, double> adj:graph[node]) {
                int next = adj.first;
                double d = adj.second;

                d += dist[node];

                if(dist[next] > d) {
                    dist[next] = d;
                    pq.push({d, {next, node}});
                }
            }
        }


        double d = dist[end_node];
        d *= -1;
        double ans = pow(2, d);

        return ans;
    }
};
70800