Submission #1295195

#	Time	Username	Problem	Language	Result	Execution time	Memory
1295195		ender_shayan	Road Construction (JOI21_road_construction)	C++20	100 / 100	1483 ms	8344 KiB

road_construction

#include <bits/stdc++.h>

using namespace std;

// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;

typedef long long ll;
typedef long double	ld;
typedef pair<ll, ll>	pii  ;
typedef pair<ll, ll>	pll  ;
typedef vector<pii>     vii  ;
typedef vector<int>     veci ;
typedef vector<pll>     vll  ;
typedef vector<ll>      vecll;

// find_by_order             order_of_key

//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define F		        first
#define S		        second
#define pb		        push_back
#define endl            '\n'
#define Mp		        make_pair
#define all(x)          x.begin(), x.end()
#define debug(x)        cerr << #x << " = " << x << endl
#define set_dec(x)	    cout << fixed << setprecision(x);
#define fast_io         ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file_io         freopen("in.txt" , "r" , stdin) ; freopen("out.txt" , "w" , stdout);
#define lb              lower_bound
#define ub              upper_bound
#define for1(n)         for(int i=1;i<=n;i++)
#define for0(n)         for(int i=0;i<n;i++)
#define forn(n)         for(int i=n;i>0;i--)
#define pq              priority_queue <pii, vector<pii>, greater<pii>>
#define int             ll

const ll mod = 1e9+7 ;// 998244353 ;// 1e9+9;

ll inf=1e18;
const int N=1e6+100,L=21,bs=701,NN=1e6;
int B[N],C[N],D[N],E[N],n,m,k,q,dp[N],pre[N],dist[N],vis[N];
vector<int>g[N];
vector<ll>ans;
pii A[N];
bool check(ll t){
    set<pll>C;
    int r=1;
    for1(n){
        while(A[r].F<=t+A[i].F && r<=n){
            auto it=C.lb({A[r].S-t,-mod*2});
            while(1){
                if(it==C.end())break;
                ll dis=max(max((ll((*it).F)-ll(A[r].S)),-(ll((*it).F)-ll(A[r].S))),max((ll((*it).S)-ll(A[r].F)),-(ll((*it).S)-ll(A[r].F))));
                if(dis>t)break;
                ans.pb(dis);
                if(ans.size()==k)return 1;
                it++;
            }
            C.insert({A[r].S,A[r].F});
            r++;
        }
        C.erase({A[i].S,A[i].F});
    }
    return 0;
}
int32_t main(){
    fast_io
    cin>>n>>k;
    for1(n){
        int x,y;cin>>x>>y;
        A[i]={x+y,x-y}; 
    }
    sort(A+1,A+n+1);
    ll l=0,r=r=mod*4;
    while(r-l!=1){
        ll mid=(l+r)>>1;
        if(check(mid)==0)l=mid;
        else r=mid;
        ans.clear();
    }
    check(l);
    sort(all(ans));
    for(int x:ans)cout<<x<<" ";cout<<endl;
    for1(k-ans.size())cout<<l+1<<endl;



}

• Subtask 1: 5 / 5
• Subtask 2: 6 / 6
• Subtask 3: 7 / 7
• Subtask 4: 20 / 20
• Subtask 5: 27 / 27
• Subtask 6: 35 / 35