<p>There is a country of <code>n</code> cities numbered from <code>0</code> to <code>n - 1</code> where <strong>all the cities are connected</strong> by bi-directional roads. The roads are represented as a 2D integer array <code>edges</code> where <code>edges[i] = [x<sub>i</sub>, y<sub>i</sub>, time<sub>i</sub>]</code> denotes a road between cities <code>x<sub>i</sub></code> and <code>y<sub>i</sub></code> that takes <code>time<sub>i</sub></code> minutes to travel. There may be multiple roads of differing travel times connecting the same two cities, but no road connects a city to itself.</p>

<p>Each time you pass through a city, you must pay a passing fee. This is represented as a <strong>0-indexed</strong> integer array <code>passingFees</code> of length <code>n</code> where <code>passingFees[j]</code> is the amount of dollars you must pay when you pass through city <code>j</code>.</p>

<p>In the beginning, you are at city <code>0</code> and want to reach city <code>n - 1</code> in <code>maxTime</code><strong> minutes or less</strong>. The <strong>cost</strong> of your journey is the <strong>summation of passing fees</strong> for each city that you passed through at some moment of your journey (<strong>including</strong> the source and destination cities).</p>

<p>Given <code>maxTime</code>, <code>edges</code>, and <code>passingFees</code>, return <em>the <strong>minimum cost</strong> to complete your journey, or </em><code>-1</code><em> if you cannot complete it within </em><code>maxTime</code><em> minutes</em>.</p>

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

<p><img alt="" src="https://assets.leetcode.com/uploads/2021/06/04/leetgraph1-1.png" style="width: 371px; height: 171px;" /></p>

<pre>
<strong>Input:</strong> maxTime = 30, edges = [[0,1,10],[1,2,10],[2,5,10],[0,3,1],[3,4,10],[4,5,15]], passingFees = [5,1,2,20,20,3]
<strong>Output:</strong> 11
<strong>Explanation:</strong> The path to take is 0 -&gt; 1 -&gt; 2 -&gt; 5, which takes 30 minutes and has $11 worth of passing fees.
</pre>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2021/06/04/copy-of-leetgraph1-1.png" style="width: 371px; height: 171px;" /></strong></p>

<pre>
<strong>Input:</strong> maxTime = 29, edges = [[0,1,10],[1,2,10],[2,5,10],[0,3,1],[3,4,10],[4,5,15]], passingFees = [5,1,2,20,20,3]
<strong>Output:</strong> 48
<strong>Explanation:</strong> The path to take is 0 -&gt; 3 -&gt; 4 -&gt; 5, which takes 26 minutes and has $48 worth of passing fees.
You cannot take path 0 -&gt; 1 -&gt; 2 -&gt; 5 since it would take too long.
</pre>

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

<pre>
<strong>Input:</strong> maxTime = 25, edges = [[0,1,10],[1,2,10],[2,5,10],[0,3,1],[3,4,10],[4,5,15]], passingFees = [5,1,2,20,20,3]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is no way to reach city 5 from city 0 within 25 minutes.
</pre>

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

<ul>
	<li><code>1 &lt;= maxTime &lt;= 1000</code></li>
	<li><code>n == passingFees.length</code></li>
	<li><code>2 &lt;= n &lt;= 1000</code></li>
	<li><code>n - 1 &lt;= edges.length &lt;= 1000</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt;= n - 1</code></li>
	<li><code>1 &lt;= time<sub>i</sub> &lt;= 1000</code></li>
	<li><code>1 &lt;= passingFees[j] &lt;= 1000</code>&nbsp;</li>
	<li>The graph may contain multiple edges between two nodes.</li>
	<li>The graph does not contain self loops.</li>
</ul>

class Solution:
    def minCost(self, maxTime: int, edges: List[List[int]], passingFees: List[int]) -> int:

        N = len(passingFees)

        # 0 -> (2, 1) -> (1, 1) -> (4, 2)

        # 8.    10.       5.        10

        # 0 -> (3, 4) -> (1, 2) -> (4, 2)
        # 8.    4.        5.        10


        # 0 -> 2, 9
        # 0 -> 2, 1
        # 0 -> 1, 9
        # 0 -> 3, 4

        # 2 -> 1, 4
        # 2 -> 0, 9
        # 2 -> 1, 9
        # 2 -> 0, 1

        # 1 -> 2, 4
        # 1 -> 0, 9
        # 1 -> 3, 2
        # 1 -> 2, 9
        # 1 -> 4, 2

        # 3 -> 0, 4
        # 3 -> 4, 10
        # 3 -> 4, 8
        # 3 -> 1, 2

    


        graph = collections.defaultdict(list)
        for u, v, dt in edges:
            if dt > maxTime: continue
            graph[u].append((dt, v))
            graph[v].append((dt, u))

        times = [math.inf]*N
        costs = [math.inf]*N
        times[0] = 0
        costs[0] = passingFees[0]

        # pq = [(costs[0], times[0], 0)]
        pq = [(costs[0], times[0], 0)]

        print(graph)


        while len(pq) > 0:
            cost, time, u = heappop(pq)
            if cost > costs[u] and time > times[u]: continue

            if u == N - 1:
                return cost

            for t, v in graph.get(u, []):
                new_t = t + time
                new_cost = cost + passingFees[v]

                if new_t > maxTime: continue

                if new_cost < costs[v]:
                    costs[v] = new_cost
                    heappush(pq, (new_cost, new_t, v))
                elif new_t < times[v]:
                    times[v] = new_t
                    heappush(pq, (new_cost, new_t, v))

        return -1

        N = len(passingFees)

        times = [float("inf")] * N
        costs = [float("inf")] * N
        times[0] = 0
        costs[0] = passingFees[0]

        pq = [(0, 0, 0)]

        graph = collections.defaultdict(set)

        for u, v, w in edges:
            graph[u].add((w, v))
            graph[v].add((w, u))



        while len(pq) > 0:
            cost, time, u = heappop(pq)
            if cost > costs[u] and time > times[u]: continue

            if u == N - 1:
                return cost + passingFees[u]

            for t, v in graph.get(u, []):
                new_t = t + time
                new_cost = cost + passingFees[u]

                if new_t > maxTime: continue

                if new_cost < costs[v]:
                    costs[v] = new_cost
                    heappush(pq, (new_cost, new_t, v))

                elif new_t < times[v]:
                    times[v] = new_t
                    heappush(pq, (new_cost, new_t, v))

        return -1
        
        
17800