Submission #1157874

#	Time	Username	Problem	Language	Result	Execution time	Memory
1157874		gazizmadi11	Road Construction (JOI21_road_construction)	C++20	6 / 100	2326 ms	30792 KiB

road_construction

//gm  --- akezhon
#include <bits/stdc++.h>
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#define pb push_back
#define pf push_front
#define F first
#define S second
#define all(v) v.begin(),v.end()
#define pii pair<int,int>
#define tm (tl+tr)/2
#define TL v+v, tl, tm
#define TR v+v+1, tm+1, tr
#define DA l <= tl && tr <= r
#define NE r < tl || tr < l
#define double long double
#define ll long long 
using namespace std;

const int N=250007;
const int mod=998244353;
const ll inf=1e10;

ll n, k;
pair<ll,ll> a[N];

ll check(int l, int r, ll d){
    if(l==r)return 0;
    int m = (l+r)/2, ans=0;
    set<int>st1, st2;
    // possible y's on each side
    for(int i=l; i <= r; i++){
        if(abs(a[m].F-a[i].F) > d)continue;
        if(i <= m)st1.insert(a[i].S);
        else st2.insert(a[i].S);
    }
    vector<int>y1,y2;
    for(int i : st1)y1.pb(i);
    for(int i : st2)y2.pb(i);
    vector<vector<int>>v1(y1.size()), v2(y2.size());
    for(int i=l; i <= r; i++){
        if(abs(a[m].F-a[i].F) > d)continue;
        if(i <= m){
            int pos = lower_bound(all(y1), a[i].S)-y1.begin();
            v1[pos].pb(a[i].F);
        }
        else{
            int pos = lower_bound(all(y2), a[i].S)-y2.begin();
            v2[pos].pb(a[i].F);    
        }
    }
    // cout << v1[0][0] << ' ' << v1[0][1] << "  " << v2[0][0] << ' ' << v2[0][1] << '\n';
    for(int i=0; i < y1.size(); i++){
        for(int j=0; j < y2.size(); j++){
            if(abs(y1[i]-y2[j]) > d)continue;
            int pos=0;
            for(int h=0; h < v1[i].size(); h++){
                while(pos < v2[j].size() && (ll)(v2[j][pos]-v1[i][h] + abs(y1[i]-y2[j])) <= d)pos++;
                ans += pos;
                if(ans > k)return inf;
            }
        }
    }
    return ans+check(l, m, d)+check(m+1, r, d);
}
vector<ll>vec;
void show(int l, int r, int d){
    if(l==r)return;
    int m = (l+r)/2, ans=0;
    set<int>st1, st2;
    // possible y's on each side
    for(int i=l; i <= r; i++){
        if(abs(a[m].F-a[i].F) > d)continue;
        if(i <= m)st1.insert(a[i].S);
        else st2.insert(a[i].S);
    }
    vector<int>y1,y2;
    for(int i : st1)y1.pb(i);
    for(int i : st2)y2.pb(i);
    vector<vector<int>>v1(y1.size()), v2(y2.size());
    for(int i=l; i <= r; i++){
        if(abs(a[m].F-a[i].F) > d)continue;
        if(i <= m){
            int pos = lower_bound(all(y1), a[i].S)-y1.begin();
            v1[pos].pb(a[i].F);
        }
        else{
            int pos = lower_bound(all(y2), a[i].S)-y2.begin();
            v2[pos].pb(a[i].F);    
        }
    }
    for(int i=0; i < y1.size(); i++){
        for(int j=0; j < y2.size(); j++){
            if(abs(y1[i]-y2[j]) > d)continue;
            for(int h=0; h < v1[i].size(); h++){
                for(int f=0; f < v2[j].size(); f++){
                    if((ll)(v2[j][f]-v1[i][h] + abs(y1[i]-y2[j])) > d)break;
                    vec.pb((ll)(v2[j][f]-v1[i][h] + abs(y1[i]-y2[j])));
                }
            }
        }
    }
    show(l, m, d);
    show(m+1, r, d);
}
void AlemAmenov(){
    cin >> n >> k;
    for(int i=1; i <= n; i++){
        cin >> a[i].F >> a[i].S;
    }
    sort(a+1, a+1+n);
    ll l=0, r=4e9, ans;
    while(l <= r){
        int m=(l+r)/2;
        if(k <= check(1, n, m))r=m-1, ans=m;
        else l=m+1;
    }
    if(check(1, n, ans) != k)ans--;
    show(1, n, ans);
    while(vec.size() != k)vec.pb(ans+1);
    sort(all(vec));
    for(int i : vec){
        cout << i << '\n';
    }
}
signed main(){

    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int RealName=1;
    // cin >> RealName;

    // freopen("newbarn.in", "r", stdin);
    // freopen("newbarn.out", "w", stdout);

    while(RealName--)
        AlemAmenov();
    
return 0;
}

• Subtask 1: 0 / 5
• Subtask 2: 6 / 6
• Subtask 3: 0 / 7
• Subtask 4: 0 / 20
• Subtask 5: 0 / 27
• Subtask 6: 0 / 35