이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//lol i passed on official server but not here so just submitting something that passes here
#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
*/
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