제출 #571592

#제출 시각아이디문제언어결과실행 시간메모리
571592zaneyu카멜레온의 사랑 (JOI20_chameleon)C++14
100 / 100
70 ms12236 KiB
#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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