제출 #677741

#제출 시각아이디문제언어결과실행 시간메모리
677741DwightKSchruteSuper Dango Maker (JOI22_dango3)C++17
7 / 100
2465 ms496 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;
}

#include "dango3.h"

void Solve(int n,int m){
    vector<bool>done(n*m+1);

    vvi who(n+1);

    for(int i=1; i<=n; i++){
        int has = m;
        vi cur(n*m);
        iota(all(cur),1);

        for(int j=1; j<=n*m && has; j++){
            if(done[j])
                continue;
            vi tmp(cur);
            tmp.erase(find(all(tmp),j));
            if(Query(tmp)==has-1){
                has--;
                cur=tmp;
                who[i].pb(j);
                done[j]=1;
            }

        }

    }

    for(int i=0; i<m; i++){
        vi cur;
        for(int j=1; j<=n; j++)
            cur.pb(who[j][i]);
        Answer(cur);
    }
}
/*
 3 2
3 3 1 2 1 2

 */
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