<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:
    vector<float> prob;
    // vector<vector<int> > adj;
    vector<vector<pair<int,float> > > adj;
    // unordered_map<string,float> m;
    priority_queue<pair<float,int> > pq;
    vector<int> visited;
    string f(int x,int y){
        return to_string(x)+to_string(y);
    }
    void bfs(int node){
        visited[node]=1;
        for(auto it:adj[node]){
            int child = it.first;
            float dist = it.second;
            // cout<<t<<" ";
            // float temp = m[f(node,child)];
            if(/*!visited[child]||*/(prob[node]*dist>prob[child])){
                // cout<<"hi";
                prob[child] = prob[node]*dist;
                pq.push({prob[child],child});
            }
            
        }
        // cout<<endl;

    }
    double maxProbability(int n, vector<vector<int>>& edges, vector<double>& succProb, int start_node, int end_node) {
        prob.resize(n);
        adj.resize(n);
        visited.resize(n);
        int i,j;

        for(i=0;i<succProb.size();i++){
            // adj[edges[i][0]].push_back(edges[i][1]);
            // adj[edges[i][1]].push_back(edges[i][0]);
            adj[edges[i][0]].push_back({edges[i][1],succProb[i]});
            adj[edges[i][1]].push_back({edges[i][0],succProb[i]});
            // m[f(edges[i][0],edges[i][1])] = succProb[i];
            // m[f(edges[i][1],edges[i][0])] = succProb[i];
            
        }
        // cout<<f(edges[0][0],edges[0][1]);
        prob[start_node] = 1;
        pq.push({1,start_node});
        // bfs(start_node);
        // for(i=0;i<n;i++)
        // cout<<prob[i]<<" ";
        // cout<<succProb.size();
        while(!pq.empty()){
            pair<float, int> k = pq.top();
            pq.pop();
            if(!visited[k.second])
            bfs(k.second);
        }
        return prob[end_node];

    }
};
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