直接加进了邻接表,不用判重边吗?

#include<bits/stdc++.h>
#define endl '\n'
using namespace std;
typedef long long ll;
const int N = 1e5;
vector<pair<int,int>>e [N];
bool vis[N];
int dis[N];
int ans = 1000005;
int n,m;
void init(){
	for(int i = 1; i <= n; i++){
		vis[i] = 0;
		dis[i] = 1000005;
	}
}
void dij(int r){
	for(int j = 0; j < e[r].size(); j++){
		init();
		dis[r] = 0;
		priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>>q;
		q.push({0,r});
		while(!q.empty()){
			int u = q.top().second;
			q.pop();
			if(vis[u]) continue;
			vis[u] = 1;
			for(auto i : e[u]){
				int v = i.first;
				int w = i.second;
				if(vis[v]) continue;
				if(u == r && v == e[r][j].first) continue;
				if(dis[v] > dis[u] + w){
					dis[v] = dis[u] + w;
					q.push({dis[v], v});
				}
			}
		}
		ans = min(ans, dis[e[r][j].first] + e[r][j].second);
	}
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin>>n>>m;
	for(int i = 0; i < m; i++){
		int u, v, d;
		cin>>u>>v>>d;
		if(u == v) continue;
		e[u].push_back({v,d});
		e[v].push_back({u,d}); //这里
	}
	for(int i = 1; i <= n; i++){
		dij(i);
	}
	if(ans != 1000005) cout<<ans<<endl;
	else cout<<"No solution."<<endl;
	return 0;
}