Submission #677445

#	Time	Username	Problem	Language	Result	Execution time	Memory
677445		DwightKSchrute	Team Contest (JOI22_team)	C++17	17 / 100	2094 ms	17624 KiB

team

/*
#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;
    cin >> n;
    vvi a(n,vi(3));
    for(int i=0; i<n; i++){
        read(a[i]);
    }
    sort(all(a));
    a=UNQ(a);
    n=a.size();
    
    int ans=-1;
    for(int i=0; i<n; i++)
        for(int j=0; j<n; j++)
            for(int k=0; k<n; k++){
                if(a[i][0]>max(a[j][0],a[k][0]) && a[j][1]>max(a[i][1],a[k][1]) && a[k][2]>max(a[i][2],a[j][2]))
                    ans=max(ans,a[i][0]+a[j][1]+a[k][2]);
            }
    out(ans)
}

• Subtask 1: 8 / 8
• Subtask 2: 0 / 29
• Subtask 3: 9 / 9
• Subtask 4: 0 / 9
• Subtask 5: 0 / 9
• Subtask 6: 0 / 9
• Subtask 7: 0 / 27