Use priority_queue (and don't erase old entries, just skip them later) instead of set for Q.

you can use priority_queue instead. std::set is too slow.

Priority queue can be a bit faster but using set and erasing elements is fine too, I always do it.

Switch order in pair (vertex, dist) to (dist, vertex). Your current solution sorts by vertex id. This is quadratic or maybe even exponential. And don't forget about long long and bigger inf because distances can get big.

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
typedef vector<pii> vii;
typedef vector<ll> vi;
#define inf ll(1e16)

vii G[100500];
vi D,P;

void Dijkstra(ll s,ll N){
	D.assign(N+1,inf);
	P.assign(N+1,-1);
	D[s]=0;
	priority_queue<pii> Q;
	Q.push({0,s});
	while(!Q.empty()){
		auto tp=Q.top();
		Q.pop();
		ll u=tp.second;
		ll weight = -tp.first;
		if(weight>D[u]) continue;
		for(auto next:G[u]){
			ll v=next.first,wt=next.second;
			if(D[v]>weight+wt){
				P[v]=u;
				D[v]=weight+wt;
				Q.push({-D[v],v});
			}
		}
	}
}
void print_shortest_path(ll to){
	vi path;
	for(ll i=to;i!=-1;i=P[i])path.push_back(i);
	reverse(path.begin(),path.end());
	for(auto i:path)cout<<i<<" ";
}
int main(){
	ll N,m;cin>>N>>m;
	for(ll i=0;i<m;i++){
		ll a,b,w;cin>>a>>b>>w;
		if (a == b) continue;
		G[a].push_back({b,w});
		G[b].push_back({a,w});
	}
	Dijkstra(1,N);
	if(D[N]==inf)cout<<-1;
	else
		print_shortest_path(N);
}

There is your code. But slightly modified.

I think your code has a bug (error).