<p>You are given an <strong>undirected weighted</strong> <strong>connected</strong> graph containing <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>, and an integer array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>, w<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> with weight <code>w<sub>i</sub></code>.</p>

<p>Some edges have a weight of <code>-1</code> (<code>w<sub>i</sub> = -1</code>), while others have a <strong>positive</strong> weight (<code>w<sub>i</sub> &gt; 0</code>).</p>

<p>Your task is to modify <strong>all edges</strong> with a weight of <code>-1</code> by assigning them <strong>positive integer values </strong>in the range <code>[1, 2 * 10<sup>9</sup>]</code> so that the <strong>shortest distance</strong> between the nodes <code>source</code> and <code>destination</code> becomes equal to an integer <code>target</code>. If there are <strong>multiple</strong> <strong>modifications</strong> that make the shortest distance between <code>source</code> and <code>destination</code> equal to <code>target</code>, any of them will be considered correct.</p>

<p>Return <em>an array containing all edges (even unmodified ones) in any order if it is possible to make the shortest distance from </em><code>source</code><em> to </em><code>destination</code><em> equal to </em><code>target</code><em>, or an <strong>empty array</strong> if it&#39;s impossible.</em></p>

<p><strong>Note:</strong> You are not allowed to modify the weights of edges with initial positive weights.</p>

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

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/18/graph.png" style="width: 300px; height: 300px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 5, edges = [[4,1,-1],[2,0,-1],[0,3,-1],[4,3,-1]], source = 0, destination = 1, target = 5
<strong>Output:</strong> [[4,1,1],[2,0,1],[0,3,3],[4,3,1]]
<strong>Explanation:</strong> The graph above shows a possible modification to the edges, making the distance from 0 to 1 equal to 5.
</pre>

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

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/18/graph-2.png" style="width: 300px; height: 300px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 3, edges = [[0,1,-1],[0,2,5]], source = 0, destination = 2, target = 6
<strong>Output:</strong> []
<strong>Explanation:</strong> The graph above contains the initial edges. It is not possible to make the distance from 0 to 2 equal to 6 by modifying the edge with weight -1. So, an empty array is returned.
</pre>

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

<p><strong class="example"><img alt="" src="https://assets.leetcode.com/uploads/2023/04/19/graph-3.png" style="width: 300px; height: 300px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 4, edges = [[1,0,4],[1,2,3],[2,3,5],[0,3,-1]], source = 0, destination = 2, target = 6
<strong>Output:</strong> [[1,0,4],[1,2,3],[2,3,5],[0,3,1]]
<strong>Explanation:</strong> The graph above shows a modified graph having the shortest distance from 0 to 2 as 6.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 100</code></li>
	<li><code><font face="monospace">1 &lt;= edges.length &lt;= n * (n - 1) / 2</font></code></li>
	<li><code>edges[i].length == 3</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i&nbsp;</sub>&lt;&nbsp;n</code></li>
	<li><code><font face="monospace">w<sub>i</sub>&nbsp;= -1&nbsp;</font></code>or <code><font face="monospace">1 &lt;= w<sub>i&nbsp;</sub>&lt;= 10<sup><span style="font-size: 10.8333px;">7</span></sup></font></code></li>
	<li><code>a<sub>i&nbsp;</sub>!=&nbsp;b<sub>i</sub></code></li>
	<li><code>0 &lt;= source, destination &lt; n</code></li>
	<li><code>source != destination</code></li>
	<li><code><font face="monospace">1 &lt;= target &lt;= 10<sup>9</sup></font></code></li>
	<li>The graph is connected, and there are no self-loops or repeated edges</li>
</ul>

class Solution:
    def modifiedGraphEdges(self, n: int, edges: List[List[int]], source: int, destination: int, target: int) -> List[List[int]]:
        cped=[]
        cnt=0
        ind=0
        mp={}
        mp1={}
        
        for s1  in range(len(edges)):
            e=edges[s1]
            ind=s1
         
            if e[-1]==-1:
                continue
            if e[0] in mp:
                mp[e[0]].append([e[1],e[2],ind])
            else:
                mp[e[0]]=[[e[1],e[2],ind]]
            if e[1] in mp:
                mp[e[1]].append([e[0],e[2],ind])
            else:
                mp[e[1]]=[[e[0],e[2],ind]]
        
        
        shrt=0
        hpq=[]
        vs={}
        fhp=[]
        vs1={}
        inv=0
        inv=999999999999999999
        fhp.append([0,source,[]])
        
        while fhp:
            cv=heapq.heappop(fhp)
            if cv[0]>target:
                break
            if cv[1]==destination:
                print("rtc",cv[1],cv[0])
                if cv[0]<target:
                    inv=cv[0]
                elif cv[0]==target:
                    inv=target
                    
                break
            if cv[1] in vs1:
                if vs1[cv[1]]<=cv[0]:
                    continue
            vs1[cv[1]]=cv[0]
            cst=cv[0]
            if cv[1] not in mp:
                continue
            for el in mp[cv[1]]:
                    if el[0] not in vs1:
                        heapq.heappush(fhp,[cst+el[1],el[0]])
                    elif el[0] in vs1 and el[1]+cst<vs1[el[0]]:
                        heapq.heappush(fhp,[cst+el[1],el[0]])
        ans=[]
      
        print("inv",inv,target)
        if inv<target:
          
            return ans
        
        elif inv==target:
            for e in edges:
                if e[-1]==-1:
                    v=[]
                    for el in e:
                        v.append(el)
                    v[-1]=target*2
                    ans.append(v)
                    continue
                ans.append(e)
            return ans
        def rndj(source,mp,destination):
            cst=99999999999999999
            vs={}
            hpq1=[]
            hpq1.append([0,source,[]])
            vstd=[]
            while hpq1:
                cl=heapq.heappop(hpq1)
                if cl[0]>target:
                    break
                if cl[1]==destination:
                    cst=cl[0]
                    vstd=cl[-1]
                    break
                if cl[1] in vs:
                    if vs[cl[1]]<=cl[0]:
                        continue
                vs[cl[1]]=cl[0]
                if cl[1] not in mp:
                    continue
                for el in mp[cl[1]]:
                    if el[0] not in vs:
                        heapq.heappush(hpq1,[cl[0]+el[1],el[0],cl[-1]+[el[2]]])
                    elif  el[0] in vs and vs[el[0]]>el[1]+cl[0]:
                        heapq.heappush(hpq1,[cl[0]+el[1],el[0],cl[-1]+[el[2]]])
            vc=collections.Counter(vstd)
            return [cst,vc]
      
        for s in range(len(edges)):
            ans=[]
            e=edges[s]
            if e[-1]==-1:
                if e[0] in mp:
                    mp[e[0]].append([e[1],1,s])
                else:
                    mp[e[0]]=[[e[1],1,s]]
                if e[1] in mp:
                    mp[e[1]].append([e[0],1,s])
                else:
                    mp[e[1]]=[[e[0],1,s]]
                
               
                
                v=rndj(source,mp,destination)
                #print("rtv",v)
                if v[0]<=target:
                    slv=collections.Counter()
                    df=target-v[0]
                    v1=[]
                    for e1 in e:
                        v1.append(e1)
                    
                    for j in range(len(edges)):
                        if j not in v[1]:
                            if edges[j][-1]==-1:
                                vl=[]
                                for el in edges[j]:
                                    vl.append(el)
                                vl[-1]=target
                                ans.append(vl)
                            else:
                                ans.append(edges[j])
                        else:
                            if j==s:
                                v1[-1]=1+df
                                ans.append(v1)
                            else:
                                if edges[j][-1]!=-1:
                                    ans.append(edges[j])
                                    continue
                                vl=[]
                                for el in edges[j]:
                                    vl.append(el)
                                vl[-1]=1
                                ans.append(vl)
                    return ans                        
                else:
                    continue
        return []
                
        
       
     
            
        
        
       
        
        
7337