Submission #58966

#TimeUsernameProblemLanguageResultExecution timeMemory
58966BenqBall Machine (BOI13_ballmachine)C++11
100 / 100
342 ms54244 KiB
#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; int N,Q, root, mn[MX], par[MX][17], key[MX], rkey[MX], p[MX], oc[MX]; vi child[MX]; set<int> lef; void dfs(int a) { mn[a] = a; for (int i: child[a]) { dfs(i); mn[a] = min(mn[a],mn[i]); } sort(all(child[a]),[](int x, int y) { return mn[x] < mn[y]; }); } int nex = 0; int dfs2(int a) { vi v; for (int i: child[a]) v.pb(dfs2(i)); key[++nex] = a; rkey[a] = nex; for (int i: v) par[i][0] = nex; // cout << nex << " " << a << "\n"; return nex; } void genTree() { dfs(root); dfs2(root); FOR(j,1,17) FOR(i,1,N+1) par[i][j] = par[par[i][j-1]][j-1]; } int ad() { int x = *lef.begin(); oc[x] = 1; lef.erase(x); return key[x]; } int rem(int x) { int cur = rkey[x], ans = 0; F0Rd(i,17) if (oc[par[cur][i]]) { cur = par[cur][i]; ans ^= 1<<i; } // cout << "AH " << x << " " << cur << "\n"; oc[cur] = 0; lef.insert(cur); return ans; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cin >> N >> Q; FOR(i,1,N+1) { cin >> p[i]; if (p[i] == 0) root = i; else child[p[i]].pb(i); } genTree(); FOR(i,1,N+1) lef.insert(i); F0R(i,Q) { int t,x; cin >> t >> x; if (t == 1) { int ans = 0; F0R(i,x) ans = ad(); cout << ans << "\n"; } else { cout << rem(x) << "\n"; } } } /* 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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