# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
892437 | vjudge1 | Team Contest (JOI22_team) | C++17 | 388 ms | 34388 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#define ll long long
#define str string
#define ins insert
#define ld long double
#define pb push_back
#define pf push_front
#define pof pop_front()
#define pob pop_back()
#define lb lower_bound
#define ub upper_bound
#define endl "\n"
#define fr first
#define sc second
#define mpa make_pair
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define sz size()
#define bc back()
#define ar array
#define vll vector<ll>
using namespace std;/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;*/
template <class _T>
bool chmin(_T &x, const _T &y){
if(x>y){
x=y;
return true;
}
return false;
}
template <class _T>
bool chmax(_T &x, const _T &y){
bool flag=false;
if (x<y){
x=y;flag|=true;
}
return flag;
}
//#define ordered_set tree<ll, null_type,less_equal<ll>, rb_tree_tag,tree_order_statistics_node_update>
void fre(string s){freopen((s+".in").c_str(),"r",stdin);freopen((s+".out").c_str(),"w",stdout);}
void start(){
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
const ll inf=2e18+7;
const ll mod=1e9+7;
const ll N=2e5+7;
const ld eps=1e-9;
void solve(){
ll i;
ll n,k;
cin>>n;
ll xx[n],yy[n],zz[n];
set<pair<ll,ll>> x,y,z;
for(i=0;i<n;i++){
ll a,b,c;
cin>>a>>b>>c;
xx[i]=a;
yy[i]=b;
zz[i]=c;
x.ins({a,i});
y.ins({b,i});
z.ins({c,i});
}
for(i=0;i<n;i++){
vector<ll> v;
v.pb(x.rbegin()->sc);
v.pb(y.rbegin()->sc);
v.pb(z.rbegin()->sc);
if(yy[v[1]]==yy[v[0]] || zz[v[0]]==zz[v[2]]){
x.erase(x.find({xx[v[0]],v[0]}));
y.erase(y.find({yy[v[0]],v[0]}));
z.erase(z.find({zz[v[0]],v[0]}));
}else if(xx[v[1]]==xx[v[0]] || zz[v[1]]==zz[v[2]]){
x.erase(x.find({xx[v[1]],v[1]}));
y.erase(y.find({yy[v[1]],v[1]}));
z.erase(z.find({zz[v[1]],v[1]}));
}else if(xx[v[2]]==xx[v[0]] || yy[v[1]]==yy[v[2]]){
x.erase(x.find({xx[v[2]],v[2]}));
y.erase(y.find({yy[v[2]],v[2]}));
z.erase(z.find({zz[v[2]],v[2]}));
}else{
cout<<xx[v[0]]+yy[v[1]]+zz[v[2]]<<endl;
return;
}
}
cout<<-1<<endl;
}
signed main(){
start();
ll t=1;
//cin>>t;
while(t--) solve();
return 0;
}
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