Auto comment: topic has been updated by acash (previous revision, new revision, compare).

make two identical graph G(u,v) and G'(u,v) such that G'(u,v) initialize to G(u,v) now for edge E(u,v) make and from node u to v with weight w/2 where u belongs to G and v belongs to G' now apply dijkstra's shortest path algorithm from node 0 and find distance of node n-1 of graph G' here is code for this problem

#include<bits/stdc++.h>
using namespace std;
#define ll long long

int main(){
    int n,m;
    cin>>n>>m;
    vector<vector<pair<ll,ll>>>v(2*n);
    ll x,y;
    ll c;
    for(int i=0; i<m; i++){
        cin>>x>>y>>c;
        x--;
        y--;
        v[x].push_back({c,y});
        v[x].push_back({c/2,y+n});
        v[x+n].push_back({c,y+n});
    }
    set<pair<ll,ll>>s;
    s.insert({0,0});
    vector<ll>dist(2*n,1e15);
    dist[0]=0;
    while(!s.empty()){
        auto p=s.begin();
        ll dis=p->first;
        ll node=p->second;
        s.erase(p);
        for(auto pr: v[node]){
            if(dist[pr.second]>dis+pr.first){
                auto itr=s.find({dist[pr.second],pr.second});
                if(itr!=s.end()){
                    s.erase(itr);
                }
                dist[pr.second]=dis+pr.first;
                s.insert({dis+pr.first,pr.second});
            }
        }
    }
    cout<<dist[2*n-1];
}

k is not required inside log as far I understand your solution

See this Click