1887B - Time Travel - CodeForces Solution

//ओ३म् ( ॐ) ॥ श्री गणेशाय नमः ॥  जय श्री राम ॥ ॐ नमः शिवाय

#include <bits/stdc++.h>
#include <iostream>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define pb(n) push_back(n)
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define coutyes cout<<"YES"<<endl
#define coutno cout<<"NO"<<endl
#define mp make_pair
#define ff first
#define ss second
#define rep(i, n)  for(int i=0;i<n;i++)
#define vep(i, j, n)  for(int i=j;i<n;i++)
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)

using namespace std;
using namespace __gnu_pbds;

#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
typedef tree<pair<int, int>, null_type,
        less<pair<int, int>>, rb_tree_tag,
        tree_order_statistics_node_update>
        ordered_set_pair;
// order_of_key (k) : Number of items strictly smaller than k.
// find_by_order(k) : K-th element in a set (counting from zero).
typedef long long ll;
typedef long double ld;

ll N = 1000000007;

void init_code() {
    fast_io;
#ifndef ONLINE_JUDGE
    freopen("Input1.txt", "r", stdin);
    freopen("Output1.txt", "w", stdout);
#endif
}
//2d prefix{
//to cal prefix ->pre[i][j]= v[i][j]+pre[i][j-1]+pre[i-1][j]-pre[i-1][j-1];
//to get ans from a,b->c,d ->pre[c][d]-pre[a-1][d]-pre[c][b-1]+pre[a-1][b-1];
//}

ll powe(ll a, ll n) {

    ll r = 1;
    while (n) {
        if (n % 2) {
            r = ((r % N) * (a % N)) % N;
            n--;
        }
        else {
            a = ((a % N) * (a % N)) % N;
            n = n / 2;
        }
    }
    return r;
}
int M = 1000000;
int sieve[1000001];

void createSieve() {
    for (int i = 2; i <= M; i++) {
        sieve[i] = 1;
    }
    for (int i = 2; i * i <= M; i++) {
        if (sieve[i] == 1) {
            if (i == 2) {
                for (int j = i * i; j <= M; j += i) {
                    sieve[j] = 2;
                }
            }
            else {
                for (int j = i * i; j <= M; j += 2 * i) {
                    if (sieve[j] == 1) {
                        sieve[j] = i;
                    }
                }
            }
        }
    }
}
// <-------------------KMP Algo for pattern searching------------------>
//  For KMP call MAtch() function.

vector<int> KMP(string& s) {
    int n = s.size();
    vector<int> lps(n);
    lps[0] = 0;
    for (int i = 1, j = 0; i < n; ++i) {
        while (j > 0 && s[i] != s[j])
            j = lps[j - 1];
        if (s[i] == s[j])
            ++j;
        lps[i] = j;
    }
    return lps;
}

vector<int> Match(string& s, string& t) {
    int n = s.size();
    auto lps = KMP(s);

    int m = t.size();
    vector<int> pos;
    for (int i = 0, j = 0; i < m; ++i) {
        while (j == n || (j > 0 && t[i] != s[j]))
            j = lps[j - 1];
        if (s[j] == t[i])
            ++j;

        if (j == n) {
            pos.push_back(i - n + 1);
        }
    }

    return pos;
}
// <-------------------------------------------------------->

// <--------4 term recurrence for Triangle of Mahonian numbers------->

// (A):f(n,k)=f(n−1,k)+f(n−1,k−1)+f(n−1,k−2)+⋯+f(n−1,k−n+1)
// (B):f(n,k−1)=f(n−1,k−1)+f(n−1,k−2)+f(n−1,k−3)+⋯+f(n−1,k−n+1)+f(n−1,k−n)
// (A)-(B) follows:
//  IMPORTANT:
// ⟹f(n,k)=f(n−1,k)+f(n,k−1)−f(n−1,k−n)

// <----------------------------------------------------------------------------------->

int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};

class DisjointSet {
    vector<ll> rank, parent, size;
public:
    DisjointSet(ll n) {
        rank.resize(n + 1, 0);
        parent.resize(n + 1);
        size.resize(n + 1);
        for (int i = 0; i <= n; i++) {
            parent[i] = i; size[i] = 1;
        }
    }

    int findUpar(int node) {
        if (node == parent[node])return node;
        return parent[node] = findUpar(parent[node]);
    }

    void unionByRank(int u, int v) {
        int up_u = findUpar(u);
        int up_v = findUpar(v);
        if (up_u == up_v)return;
        if (rank[up_u] < rank[up_v]) {
            parent[up_u] = up_v;
        }
        else if (rank[up_v] < rank[up_u]) {
            parent[up_v] = up_u;
        }
        else {
            parent[up_v] = up_u;
            rank[up_u]++;
        }
    }
    void unionBySize(int u, int v) {
        int up_u = findUpar(u);
        int up_v = findUpar(v);
        if (up_u == up_v)return;
        if (size[up_u] < size[up_v]) {
            parent[up_u] = up_v;
            size[up_v] += size[up_u];
        }
        else {
            parent[up_v] = up_u;
            size[up_u] += size[up_v];
        }
    }
};

// Binary Lifting to find Kth ancestor
ll up[200010][25];
vector<ll> depth(200010);
void binary_lifting(ll src, ll par, vector<vector<ll>> &adj)
{
    up[src][0] = par;
    for (int i = 1; i < 22; i++)
    {
        if (par != -1)depth[src] = depth[par] + 1;
        if (up[src][i - 1] != -1)
            up[src][i] = up[up[src][i - 1]][i - 1];
        else up[src][i] = -1;
    }

    for (int child : adj[src])
    {
        if (child != par && child != src)
            binary_lifting(child, src, adj);
    }
}
ll getKthAncestor(ll node, ll k) {
    if (depth[node] < k) {
        return -1;
    }
    for (int j = 20; j >= 0; j--) {
        if (k >= (1 << j)) {
            node = up[node][j];
            k -= 1 << j;
        }
    }
    return node;
}

void solve() {
    ll n, k, ans = 0; cin >> n >> k;
    vector<vector<ll>> adj(n + 1);
    map<pair<ll, ll>, vector<ll>> mp;
    rep(i, k) {
        ll m; cin >> m;
        rep(j, m) {
            ll u, v; cin >> u >> v;
            adj[u].pb(v); adj[v].pb(u);
            mp[ {u, v}].pb(i + 1);
            mp[ {v, u}].pb(i + 1);
        }
    }
    ll t; cin >> t; vector<ll> a(t);
    vector<vector<ll>> tt(k + 1);
    rep(i, t) {
        cin >> a[i];
        tt[a[i]].pb(i + 1);
    }
    vector<ll> dis(n + 1, 1e18), vis(n + 1);
    priority_queue<pair<ll, ll>> pq;
    pq.push({0, 1}); dis[1] = 0;
    while (!pq.empty()) {
        auto it = pq.top(); pq.pop();
        ll d = -1 * it.first, nd = it.second;
        if (vis[nd] == 1)continue;
        vis[nd] = 1;
        for (auto j : adj[nd]) {
            if (vis[j] == 0) {
                ll mn = 1e18;
                for (auto i : mp[ {nd, j}]) {
                    ll jj = upper_bound(all(tt[i]), d) - tt[i].begin();
                    if (jj != tt[i].size()) {
                        jj = tt[i][jj];
                        mn = min(mn, jj);
                    }
                }
                if (dis[j] > mn) {
                    dis[j] = mn;
                    pq.push({ -mn, j});
                }
            }
        }
    }
    ans = dis[n];
    if (ans == 1e18)ans = -1;
    cout << ans << endl;

}

int main() {
    init_code();

    ll t;
    t = 1;
    // cin >> t;
    while (t--) {

        solve();
    }

    return 0;
}