Submission #1124976

#	Time	Username	Problem	Language	Result	Execution time	Memory
1124976		acoatoitgs	Bitaro’s Party (JOI18_bitaro)	C++17	0 / 100	2094 ms	584 KiB

bitaro

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;

#define ll long long
#define pb push_back
#define m_pi 2 * acos(0.0)
#define all(a) (a).begin(), (a).end()
#define LL_INF 0x3f3f3f3f3f3f3f3f
#define INF 0x3f3f3f3f
void solve();

constexpr bool isTc = 0;
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    if (isTc)
    {
        int T;
        cin >> T;
        while (T--)
        {
            solve();
        }
    }
    else
        solve();
    return 0;
}

/*######################################*/

int SQRT;

vector<vector<int>> adj, radj;
int N, M, Q;

void i_merge(vector<pair<int,int>> &S, vector<pair<int,int>> &T)
{
    
}
void precompute(vector<vector<pair<int,int>>> &pre)
{
    for(int i = 0; i <= N; i++)
    {
        pre[i].push_back({i, 0});
    }

    for(int i = 1; i <= N; i++)
    {
        for(auto edge : adj[i])
        {
            i_merge(pre[i], pre[edge]);
        }
    }
}

int dp(int node, bitset<100005> &bs)
{
    int best = 0;    
    for(auto i : radj[node])
    {   
        int d = dp(i, bs);
        if(d != 0 || !bs[i])
            best = max(best, d+1);
    }

    return best;
}

void solve()
{
    cin >> N >> M >> Q;

    adj.resize(N+1);
    radj.resize(N+1);
    for(int i = 0; i < M; i++)
    {
        int a,b;
        cin >> a >> b;
        adj[a].emplace_back(b);
        radj[b].emplace_back(a);
    }
    SQRT = sqrt(N);
    
    vector<vector<pair<int,int>>> pre(N+1);
    precompute(pre);

    bitset<100005> rm;
    while(Q--)
    {   
        int T, Y, x;
        vector<int> m;
        cin >> T >> Y;
        m.reserve(Y);
        
        for(int i = 0; i < Y; i++)
        {
            cin >> x;
            rm[x] = 1;
            m.push_back(x);
        }

        int ans;
        if(true)
        {   
            ans = dp(T, rm);
        }
        else
        {

        }


        cout << (ans == 0 && rm[T] ? -1 : ans) << "\n";
        for(auto i : m) rm[i] = 0;
    }
}

• Subtask 1: 0 / 7
• Subtask 2: 0 / 7
• Subtask 3: 0 / 86