제출 #47805

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
47805		qoo2p5	Bitaro’s Party (JOI18_bitaro)	C++17	14 / 100	1004 ms	247116 KiB

bitaro

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int INF = (int) 1e9 + 1e6 + 123;
const ll LINF = (ll) 1e18 + 1e9 + 123;

#define rep(i, s, t) for (auto i = (s); i < (t); ++(i))
#define per(i, s, t) for (auto i = (s); i >= (t); --(i))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define mp make_pair
#define pb push_back

bool mini(auto &x, const auto &y) {
    if (y < x) {
        x = y;
        return 1;
    }
    return 0;
}

bool maxi(auto &x, const auto &y) {
    if (y > x) {
        x = y;
        return 1;
    }
    return 0;
}

void run();

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    run();
    return 0;
}

const int N = (int) 1e5 + 123;
const int K = 300;

int n, m, q;
vector<int> g[N];
int ans[N];
int t[N], y[N];
vector<int> c[N];

int dp[N];

int naive(int id) {
    fill(dp, dp + N, 0);
    for (int v : c[id]) {
        dp[v] = -INF;
    }
    rep(i, 1, n + 1) {
        for (int j : g[i]) {
            if (dp[j] != -INF) {
                maxi(dp[i], dp[j] + 1);
            }
        }
    }
    return (dp[t[id]] == -INF ? -1 : dp[t[id]]);
}

bool used[N];

int cnt[N];
pair<int, int> best[N][K];
pair<int, int> tmp[2 * K];

int small(int id) {
    for (int v : c[id]) {
        used[v] = 1;
    }
    
    int res = -INF;
    
    rep(i, 0, cnt[t[id]]) {
        auto &it = best[t[id]][i];
        if (!used[it.second]) {
            maxi(res, it.first);
        }
    }
    
    for (int v : c[id]) {
        used[v] = 0;
    }
    
    return (res == -INF ? -1 : res);
}

void run() {
    cin >> n >> m >> q;
    rep(i, 0, m) {
        int u, v;
        cin >> u >> v;
        g[v].pb(u);
    }
    
    rep(i, 0, N) {
        rep(j, 0, K) {
            cnt[i] = 0;
            best[i][j] = {-INF, -1};
        }
    }
    
    rep(i, 1, n + 1) {
        cnt[i] = 1;
        best[i][0] = {0, i};
        
        for (int j : g[i]) {
            rep(t, 0, cnt[j]) {
                best[j][t].first++;
            }
            merge(best[i], best[i] + cnt[i], best[j], best[j] + cnt[j], tmp);
            rep(t, 0, cnt[j]) {
                best[j][t].first--;
            }
            int tot = cnt[i] + cnt[j];
            cnt[i] = min(tot, K);
            int id = 1;
            per(t, cnt[i] - 1, 0) {
                best[i][t] = tmp[tot - id];
                id++;
            }
        }
    }
    
    rep(i, 0, q) {
        cin >> t[i] >> y[i];
        c[i].resize(y[i]);
        rep(j, 0, y[i]) {
            cin >> c[i][j];
        }
        
        if (y[i] >= K) {
            ans[i] = naive(i);
        } else {
            ans[i] = small(i);
        }
    }
    
    rep(i, 0, q) {
        cout << ans[i] << "\n";
    }
}

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