Maintaining two arrays of shortest distances from node 1 and n respectively using Dijkstra's Algorithm. And then go through all the edges, applying coupon on each one and saving the least cost.

#include <bits/stdc++.h>

using namespace std;

#define ar array
#define ll long long

const int MAX_N = 1e5 + 1;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const ll LINF = 1e18;
const int sz = 1e5+1;

int n,m;
vector<bool> prd(sz);
void dijkstra(vector<vector<pair<int, int> > >& adj, vector<ll>& dist, int k)
{
    for(int i=1; i<sz; ++i)
    {
        prd[i]=false;
    }
    priority_queue<pair<int, int> > q;
    dist[k]=0;
    q.push({0,k});
    while(!q.empty())
    {
        int a = q.top().second;
        q.pop();
        if(prd[a])continue;
        prd[a]=true;
        for(auto u : adj[a])
        {
            int b = u.first, w = u.second;
            if(dist[a]+w<dist[b])
            {
                dist[b]=w+dist[a];
                q.push({-dist[b],b});
            }
        }
    }
}
void solve()
{
    cin>>n>>m;
    vector<vector<pair<int, int> > > adj(sz), adj1(sz);
    vector<tuple<int, int, int> > edges;
    for(int i=0; i<m; ++i)
    {
        int a,b,w;
        cin>>a>>b>>w;
        adj[a].push_back({b,w});
        adj1[b].push_back({a,w});
        edges.push_back({a,b,w});
    }
    vector<ll> dist(sz, LINF), dist1(sz,LINF);
    dijkstra(adj, dist,1);
    dijkstra(adj1, dist1,n);
    ll ans = LINF;
    for(auto e : edges)
    {
        int a,b,w;
        tie(a,b,w) = e;
        ans = min(ans, dist[a]+dist1[b]+w/2);
    }
    cout<<ans<<endl;
}

int main()
{
    solve();
}