Submission #61423

#	Time	Username	Problem	Language	Result	Execution time	Memory
61423		Benq	Bitaro’s Party (JOI18_bitaro)	C++11	100 / 100	1972 ms	411048 KiB

bitaro

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

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

const int BLOCK = 300;

int N,M,Q, BES[MX];
vpi bes[MX];
bitset<MX> bad;
vi in[MX];

vpi merge(const vpi& a, const vpi& b) {
    vpi c;
    int i0 = 0, i1 = 0;
    while (sz(c) < BLOCK) {
        if (i0 < sz(a)) {
            if (i1 < sz(b)) {
                if (a[i0] > b[i1]) {
                    if (!bad[a[i0].s]) bad[a[i0].s] = 1, c.pb(a[i0]);
                    i0 ++;
                } else {
                    if (!bad[b[i1].s]) bad[b[i1].s] = 1, c.pb(b[i1]);
                    i1 ++;
                }
            } else {
                if (!bad[a[i0].s]) bad[a[i0].s] = 1, c.pb(a[i0]);
                i0 ++;
            }
        } else {
            if (i1 < sz(b)) {
                if (!bad[b[i1].s]) bad[b[i1].s] = 1, c.pb(b[i1]);
                i1 ++;
            } else {
                break;
            }
            
        }
    }
    for (auto x: c) bad[x.s] = 0;
    return c;
}

void gen(int x) {
    bes[x].pb({0,x});
    for (int i: in[x]) {
        vpi v = bes[i]; for (auto& a: v) a.f ++;
        bes[x] = merge(bes[x],v);
    }
}

void input() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> M >> Q;
    F0R(i,M) {
        int S,E; cin >> S >> E;
        in[E].pb(S);
    }
    FOR(i,1,N+1) gen(i);
}

void GEN(int x) {
    if (bad[x]) BES[x] = -MOD;
    else BES[x] = 0;
    for (int i: in[x]) BES[x] = max(BES[x],BES[i]+1);
}

int smart(int T) {
    for (auto a: bes[T]) if (!bad[a.s]) return a.f;
    return -1;
}

int brute(int T) {
    FOR(i,1,T+1) GEN(i);
    return max(BES[T],-1);
}

int main() {
    input();
    F0R(i,Q) {
        int T,Y; cin >> T >> Y;
        vi v(Y); F0R(j,Y) cin >> v[j];
        for (int i: v) bad[i] = 1;
        // cout << brute(T) << "\n";
        if (Y >= BLOCK) cout << brute(T) << "\n";
        else cout << smart(T) << "\n";
        for (int i: v) bad[i] = 0;
    }
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

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