이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*
#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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