Submission #903084

#TimeUsernameProblemLanguageResultExecution timeMemory
903084GrindMachineSynchronization (JOI13_synchronization)C++17
40 / 100
8080 ms22624 KiB
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

vector<pll> adj[N];
vector<pll> times[N];
ll ans;

void dfs1(ll u, ll p, ll t){
    ans++;

    for(auto [v,id] : adj[u]){
        if(v == p) conts;
        auto &ti = times[id];

        // find the biggest l s.t l <= t
        auto it = upper_bound(all(ti),make_pair(t+1,-1ll));
        if(it == ti.begin()) conts;

        it--;
        auto [l,r] = *it;
        dfs1(v,u,min(t,r));
    }
}

void solve(int test_case)
{
    ll n,m,q; cin >> n >> m >> q;
    vector<pll> edges(n+5);

    rep1(i,n-1){
        ll u,v; cin >> u >> v;
        adj[u].pb({v,i}), adj[v].pb({u,i});
        edges[i] = {u,v};
    }

    vector<bool> curr(n+5);

    rep1(i,m){
        ll id; cin >> id;
        if(!curr[id]){
            curr[id] = 1;
            times[id].pb({i,m});
        }
        else{
            curr[id] = 0;
            times[id].back().ss = i;
        }
    }

    while(q--){
        ll root; cin >> root;
        ans = 0;
        dfs1(root,-1,m+1);
        cout << ans << endl;
    }
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}
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