제출 #676671

#제출 시각아이디문제언어결과실행 시간메모리
676671DwightKSchruteEvent Hopping 2 (JOI21_event2)C++17
1 / 100
44 ms8212 KiB
/* #pragma GCC target ("avx2") #pragma GCC optimization ("O3") #pragma GCC optimization ("unroll-loops") */ #include<bits/stdc++.h> //#include <ext/pb_ds/assoc_container.hpp> //#include <ext/pb_ds/tree_policy.hpp> //using namespace __gnu_pbds; using namespace std; typedef long double ld; typedef long long ll; typedef unsigned long long ull; typedef vector<int>vi; typedef vector<vector<int>>vvi; typedef vector<ll>vl; typedef vector<vl> vvl; typedef pair<int,int>pi; typedef pair<ll,ll> pl; typedef vector<pl> vpl; typedef vector<ld> vld; typedef pair<ld,ld> pld; typedef vector<pi> vpi; //typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set; template<typename T> ostream& operator<<(ostream& os, vector<T>& a){os<<"[";for(int i=0; i<ll(a.size()); i++){os << a[i] << ((i!=ll(a.size()-1)?" ":""));}os << "]\n"; return os;} #define all(x) x.begin(),x.end() #define YES out("YES") #define NO out("NO") #define out(x){cout << x << "\n"; return;} #define outfl(x){cout << x << endl;return;} #define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL) #define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";} #define pb push_back #define umap unordered_map template<typename T> void read(vector<T>& v){ int n=v.size(); for(int i=0; i<n; i++) cin >> v[i]; } template<typename T> vector<T>UNQ(vector<T>a){ vector<T>ans; for(T t:a) if(ans.empty() || t!=ans.back()) ans.push_back(t); return ans; } void solve(); int main(){ GLHF; int t=1; //cin >> t; while(t--) solve(); } void solve() { int n,k; cin >> n >> k; vvi a(n,vi(3)); for(int i=0; i<n ;i++){ cin >> a[i][0] >> a[i][1]; a[i][1]--; a[i][2]=i; } sort(all(a)); vi ans(k,1e9); for(int i=0; i<(1<<n); i++){ if(__builtin_popcount(i)!=k) continue; vi cur; for(int j=0; j<n; j++) if(i&(1<<j)) cur.pb(j); bool bad=0; for(int j=0; j+1<k; j++) if(a[cur[j]][1] >= a[cur[j+1]][0]) bad=1; if(bad) continue; for(int j=0; j<k; j++) cur[j] = 1+a[cur[j]][2]; sort(all(cur)); ans=min(ans,cur); } if(ans[0]>n)out(-1) for(int x:ans) cout <<x << "\n"; }
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