이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
#define FAST_IO ios_base::sync_with_stdio(0); cin.tie(0)
#define FOR(i, a, b) for(int i = (a); i <= (b); i++)
#define REP(n) FOR(O, 1, (n))
#define pb push_back
#define f first
#define s second
typedef long double ld;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<pii> vii;
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MAXN = 100100, MAXK = 40;
const int INF = 1e9;
//const ll MOD = 1e9+7;
//const ll MOD = 998244353;
const int C = 300;
//const int C = 0;
int n, m, q;
vi adj[MAXN];
bool busy[MAXN];
vi busySeq;
int dp[MAXN];
vii pre[MAXN];
//priority_queue<pii, vii, greater<pii>> pq[MAXN];
//vii tmp;
/*void clean (int id) {
while (pq[id].size() > C) {
pq[id].pop();
}
}
void save (int id) {
while (!pq[id].empty()) {
auto p = pq[id].top();
pre[id].pb(p);
pq[id].pop();
}
reverse(pre[id].begin(), pre[id].end());
}*/
vii tmp;
bool added[MAXN];
vi tmpAdded;
void pass (int a, int b) {
tmp.clear();
tmpAdded.clear();
int id1 = 0, id2 = 0;
while (id1 < (int)pre[a].size() || id2 < (int)pre[b].size()) {
if ((int)tmp.size() > C) break;
if ( (id2 >= (int)pre[b].size()) ||
(id1 < (int)pre[a].size() && pre[a][id1].f+1 > pre[b][id2].f) ) {
int from = pre[a][id1].s;
if (!added[from]) {
tmp.pb({pre[a][id1].f+1, pre[a][id1].s});
added[from] = true;
tmpAdded.pb(from);
}
id1++;
} else {
int from = pre[b][id2].s;
if (!added[from]) {
added[from] = true;
tmpAdded.pb(from);
tmp.pb(pre[b][id2]);
}
id2++;
}
}
swap(pre[b], tmp);
for (int x : tmpAdded) added[x] = false;
}
int main()
{
FAST_IO;
cin >> n >> m >> q;
REP(m) {
int s, e;
cin >> s >> e;
adj[s].pb(e);
}
FOR(i, 1, n) {
//pq[i].push({0, i});
//clean(i);
//save(i);
pre[i].pb({0, i});
for (int u : adj[i])
pass (i, u);
}
REP(q) {
int t, y;
cin >> t >> y;
busySeq.clear();
REP(y) {
int x; cin >> x;
busySeq.pb(x);
busy[x] = true;
}
if (y >= C) {
FOR(i, 1, t) dp[i] = busy[i] ? (-1) : 0;
FOR(i, 1, t) {
if (dp[i] < 0) continue;
for (int u : adj[i])
dp[u] = max(dp[u], dp[i] + 1);
}
cout << dp[t] << "\n";
} else {
int ans = -1;
for (auto p : pre[t]) {
int len = p.f, from = p.s;
if (busy[from]) continue;
ans = len;
break;
}
cout << ans << "\n";
}
for (int x : busySeq) busy[x] = false;
}
return 0;
}
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