| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 990288 | VinhLuu | 팀들 (IOI15_teams) | C++17 | 0 ms | 0 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#include <bits/stdc++.h>
#define ll long long
#define fi first
#define se second
#define pb push_back
#define all(lmao) lmao.begin(), lmao.end()
using namespace std;
typedef pair<int,int> pii;
typedef tuple<int,int,int> tp;
const int N = 5e5 + 5;
int block = 555;
const int mod = 1e9 + 7;
//const ll oo = 5e18;
int n, cnt, R[N], m;
struct NODE{
int l, r, w;
} st[50000000];
int Newchild(int val){
cnt++;
st[cnt].w = val;
st[cnt].l = st[cnt].r = 0;
return cnt;
}
int Newnode(int l,int r){
cnt++;
st[cnt].l = l, st[cnt].r = r;
st[cnt].w = st[l].w + st[r].w;
return cnt;
}
int build(int l,int r){
if(l == r) return Newchild(0);
int mid = (l + r) / 2;
return Newnode(build(l, mid), build(mid + 1, r));
}
int update(int i,int l,int r,int u){
if(l > u || r < u) return i;
if(l == r) return Newchild(st[i].w + 1);
int mid = (l + r) / 2;
return Newnode(update(st[i].l, l, mid, u), update(st[i].r, mid + 1, r, u));
}
int get(int i,int l,int r,int u,int v){
if(l > r || l > v || r < u) return 0;
if(u <= l && r <= v) return st[i].w;
int mid = (l + r) / 2;
return get(st[i].l, l, mid, u, v) + get(st[i].r, mid + 1, r, u, v);
}
vector<int> add[N];
int sign[N];
void init(int _n, int A[], int B[]){
n = _n;
m = 0;
vector<int> vr;
for(int i = 0; i < n; i ++) vr.pb(i);
sort(all(vr), [&] (int x,int y){return (B[x] == B[y] ? A[x] < A[y] : B[x] < B[y]);});
int ptr = 0;
for(int i = 1; i <= n + 1; i ++) sign[i] = n + 1;
for(int i = n - 1; i >= 0; i --){
// cerr << B[vr[i]] << " ";
sign[B[vr[i]]] = i + 1;
B[vr[i]] = i + 1;
}
// cerr << "\n";
for(int i = n; i >= 1; i --){
sign[i] = min(sign[i + 1], sign[i]);
}
for(int i = 0; i < n; i ++){
add[A[i]].pb(B[i]);
}
R[0] = build(1, n);
for(int i = 1; i <= n; i ++){
int cur = R[i - 1], tmp = 0;
for(auto j : add[i]){
tmp = update(cur, 1, n, j);
swap(tmp, cur);
}
R[i] = cur;
}
}
ll mark[N], sub;
vector<int> K;
vector<pii> sk;
ll cal(int i,int mid){
if(mid == 0) return 0;
ll val = get(R[K[i]], 1, n, 1, mid);
if(sk.empty()) val -= sub;
else if(sk[0].fi <= mid) val -= sub;
else{
int le = 0, ri = sk.size() - 1, mi, u = -1;
while(le <= ri){mi = (le + ri) / 2;
if(sk[mi].fi >= mid) u = mi, le = mi + 1; else ri = mi - 1;}
int lef = (u == sk.size() - 1 ? 1 : sk[u + 1].fi + 1);
val = val - (sub - mark[sk[u].se] + get(R[K[sk[u].se]], 1, n, lef, mid));
}
return val;
}
int can(int M,int k[]){
K.clear(), sk.clear();
int sum = 0;
for(int i = 0; i < M; i ++){
mark[i] = 0, K.pb(k[i]);
sum += k[i];
}
sort(all(K));
if(K.back() > n || sum > n) return 0;
sub = 0;
for(int i = 0; i < M; i ++){
int l = sign[K[i]], r = n, mid, ans = -1;
ll w = cal(i, sign[K[i]] - 1);
while(l <= r){
mid = (l + r) / 2;
ll val = cal(i, mid) - w;
if(val >= K[i]){
ans = mid;
r = mid - 1;
}else l = mid + 1;
}
if(ans == -1) return 0;
int tmp = 1; bool gg = 0;
while(!sk.empty() && sk.back().fi <= ans){
sub = sub - get(R[K[sk.back().se]], 1, n, tmp, sk.back().fi);
tmp = sk.back().fi + 1;
gg = 1;
sk.pop_back();
}
if(!sk.empty() && gg){
mark[sk.back().se] += get(R[K[sk.back().se]], 1, n, 1, tmp - 1);
sub += get(R[K[sk.back().se]], 1, n, 1, tmp - 1);
}
if(!sk.empty()){
int lmao = get(R[K[sk.back().se]], 1, n, 1, ans);
mark[sk.back().se] -= lmao;
sub -= lmao;
}
mark[i] = get(R[K[i]], 1, n, 1, ans) + (sk.empty() ? 0 : mark[sk.back().se]);
sub += get(R[K[i]], 1, n, 1, ans);
sk.pb({ans, i});
}
return 1;
}
#define lpv
#ifdef lpv
int n_, a_[N], b_[N];
int inQ;
int inM, inK[N];
signed main(){
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define task "v"
if(fopen(task ".inp","r")){
freopen(task ".inp","r",stdin);
freopen(task ".out","w",stdout);
}
cin >> n_ >> inQ;
for (int i = 0; i < n_; ++i) cin >> a_[i] >> b_[i];
init(n_, a_, b_);
while (inQ--) {
cin >> inM;
for (int i = 0; i < inM; ++i) cin >> inK[i];
int u = can(inM, inK);
cout << u << "\n";
// cerr << u;
}
// cerr << "\n";
/*
10 10
2 4
3 3
4 10
4 4
3 9
3 10
2 2
3 5
1 10
2 4
3
1 1 5
5
3 2 3 3 1
2
3 2
1
5
2
5 2
4
5 4 1 1
3
3 5 3
2
5 5
4
5 5 5 4
3
1 5 3
*/
}
#endif // lpv