<p>You are given a network of <code>n</code> nodes, labeled from <code>1</code> to <code>n</code>. You are also given <code>times</code>, a list of travel times as directed edges <code>times[i] = (u<sub>i</sub>, v<sub>i</sub>, w<sub>i</sub>)</code>, where <code>u<sub>i</sub></code> is the source node, <code>v<sub>i</sub></code> is the target node, and <code>w<sub>i</sub></code> is the time it takes for a signal to travel from source to target.</p>

<p>We will send a signal from a given node <code>k</code>. Return <em>the <strong>minimum</strong> time it takes for all the</em> <code>n</code> <em>nodes to receive the signal</em>. If it is impossible for all the <code>n</code> nodes to receive the signal, return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/23/931_example_1.png" style="width: 217px; height: 239px;" />
<pre>
<strong>Input:</strong> times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 1
<strong>Output:</strong> 1
</pre>

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

<pre>
<strong>Input:</strong> times = [[1,2,1]], n = 2, k = 2
<strong>Output:</strong> -1
</pre>

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

<ul>
	<li><code>1 &lt;= k &lt;= n &lt;= 100</code></li>
	<li><code>1 &lt;= times.length &lt;= 6000</code></li>
	<li><code>times[i].length == 3</code></li>
	<li><code>1 &lt;= u<sub>i</sub>, v<sub>i</sub> &lt;= n</code></li>
	<li><code>u<sub>i</sub> != v<sub>i</sub></code></li>
	<li><code>0 &lt;= w<sub>i</sub> &lt;= 100</code></li>
	<li>All the pairs <code>(u<sub>i</sub>, v<sub>i</sub>)</code> are <strong>unique</strong>. (i.e., no multiple edges.)</li>
</ul>

from heapq import heappush, heappop
class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        graph=defaultdict(list)
        for x in times:
            graph[x[0]].append((x[1], x[2]))
        
        print(graph)
        
        dstarr=[10**10]*n
        hp=[(0,k)]
        visited=set()
        dstarr[k-1]=0
        
        while hp:
            dst,node=heappop(hp)
            if node in visited:
                continue
            visited.add(node)

            for nbnode,nbw in graph[node]:
                dstarr[nbnode-1]=min(dstarr[nbnode-1],nbw+dstarr[node-1])
                if nbnode not in visited:
                    heappush(hp, (dstarr[nbnode-1], nbnode))
        
        res=max(dstarr)

        return res if res<10**10 else -1

        # # Build graph
        # graph={}
        # for time in times:
        #     if time[0]-1 in graph:
        #         graph[time[0]-1].append((time[1]-1, time[2]))
        #     else:
        #         graph[time[0]-1]=[(time[1]-1, time[2])]

        # res=[10**10]*n
        # q=[(0, k-1)]
        # res[k-1]=0
        # visited=set()

        # while q:
        #     dst,idx=heappop(q)

        #     if idx in visited:
        #         continue
        #     visited.add(idx)

        #     if idx in graph:
        #         for nbh in graph[idx]:
        #             distance=dst+nbh[1]
        #             res[nbh[0]]=min(res[nbh[0]], distance)
        #             if nbh[0] not in visited:
        #                 heappush(q, (res[nbh[0]],nbh[0]))
        
        # r=max(res)
        # if r==10**10:
        #     return -1
        # else:
        #     return r

            





        
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