HackerEarth Metro problem solution

In this HackerEarth Metro problem solution A city subway line has become huge and it is hard to take the shortest path through them. You have to find the shortest path in subway lines. In the second 0, you are in the station start and you want to go to the station end.

The city has  stations. The subway has m lines. Each subway line goes to some stations.

The i – th subway goes to stations u(i,1),u(i,2),u(i,3),…,u(ii,ki) (in order) and this train takes w(i,j) seconds to travel from u(i,j) to u(i,j+1)(for 1 <=i <= m and 1 <= j <= ki).

Trains are ready for the passengers to get in, but the last train goes in the second ti(and you are allowed to board it in between the path).

#include<bits/stdc++.h>
using namespace std;
#define int long long
typedef pair<int, int> pii;
typedef pair<int, pii> pi_ii;
const int maxn = 1e5;
const int inf = 1e18;

int n, m, st, en;
//edge: u v last w
vector<pi_ii> vc[maxn];

void init(){
    int M = 0;
    cin >> n >> m;
    for(int _ = 0; _ < m; _++){
  int t, last;
  cin >> t >> last;
  int v[t];
  int w[t-1];
  
  for(int j = 0; j < t; j++)
      cin >> v[j], v[j]--;

  for(int j = 0; j + 1 < t; j++)
      cin >> w[j];
  int cur_last = last;
  for(int j = 0; j + 1 < t; j++){
      vc[v[j]].push_back(pi_ii(v[j + 1], pii(w[j], cur_last)));
      cur_last += w[j];
  }
  M += t;
    }
    cin >> st >> en;
    st--;
    en--;
    m = M;
}

int dis[maxn];
void DIJKSTRA(int start){
    fill(dis, dis + n, inf);
    dis[start] = 0;
    set<pii> st;
    for(int i = 0; i < n; i++)
  st.insert(pii(dis[i], i));
    
    while(st.size()){
  int u = st.begin()->second;
  st.erase(st.begin());
  for(pi_ii edge: vc[u]){
      int v = edge.first;
      int w = edge.second.first;
      int last = edge.second.second;
      if(dis[u] <= last){
    int cur = dis[u] + w;
    if(cur < dis[v]) {
        st.erase(pii(dis[v], v));
        dis[v] = cur;
        st.insert(pii(dis[v], v));
    }
      }
  }
    }
}


main(){
    init();
    DIJKSTRA(st);
    if(dis[en] == inf)
  cout << -1 << endl;
    else
  cout << dis[en] << endl;
}