이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "chameleon.h"
#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<long long,null_type,less_equal<long long>,rb_tree_tag,tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("Ofast")
//#pragma GCC target("avx2")
//order_of_key #of elements less than x
// find_by_order kth element
using ll=long long;
using ld=long double;
using pii=pair<int,int>;
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()),c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
const ll maxn=5e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const int MOD=1e9+7;
const ld PI=acos(-1);
const ld eps=1e-6;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
template<typename T1,typename T2>
ostream& operator<<(ostream& out,pair<T1,T2> P){
out<<P.f<<' '<<P.s;
return out;
}
template<typename T>
ostream& operator<<(ostream& out,vector<T> V){
REP(i,sz(V)) out<<V[i]<<((i!=sz(V)-1)?" ":"");
return out;
}
namespace {
vector<int> vv[maxn];
bool hv(vector<int> v,int x){
v.pb(x);
return Query(v)!=sz(v);
}
int col[maxn];
void dfs(int u,int c){
col[u]=c;
for(int x:vv[u]){
if(!col[x]) dfs(x,3-c);
}
}
} // namespace
void Solve(int n) {
REP1(i,2*n){
vector<int> v[2];
REP1(j,i-1) col[j]=0;
REP1(j,i-1){
if(!col[j]) dfs(j,1);
}
REP1(j,i-1) v[col[j]-1].pb(j);
REP(j,2){
while(hv(v[j],i)){
int l=0,r=sz(v[j])-1;
while(l<r){
int mid=(l+r)/2;
vector<int> vv;
REP(k,mid+1) vv.pb(v[j][k]);
if(hv(vv,i)) r=mid;
else l=mid+1;
}
int x=v[j][l];
vv[i].pb(x),vv[x].pb(i);
v[j].erase(v[j].begin(),v[j].begin()+l+1);
}
}
}
//REP1(i,2*n) cout<<i<<' '<<vv[i]<<'\n';
REP1(i,2*n){
for(int x:vv[i]){
assert(Query({i,x})==1);
if(i>x) continue;
vector<int> vv;
REP1(k,2*n){
if(k==i or k==x) continue;
vv.pb(k);
}
if(Query(vv)<=n-1){
Answer(i,x);
break;
}
}
}
}
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