Submission #321300

#TimeUsernameProblemLanguageResultExecution timeMemory
321300zaneyuJust Long Neckties (JOI20_ho_t1)C++14
100 / 100
330 ms19556 KiB
/*input
5
4 7 9 10 11 12
3 5 7 9 11
*/
//#include "supertrees.h"
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef tree<long long, null_type, less_equal<long long>, rb_tree_tag, tree_order_statistics_node_update> indexed_set;
#pragma GCC optimize("unroll-loops,no-stack-protector")
//order_of_key #of elements less than x
// find_by_order kth element
typedef long long int ll;
#define ld double
#define pii pair<int,int>
#define f first
#define s second
#define pb push_back
#define REP(i,n) for(ll i=0;i<n;i++)
#define REP1(i,n) for(int i=1;i<=n;i++)
#define FILL(n,x) memset(n,x,sizeof(n))
#define ALL(_a) _a.begin(),_a.end()
#define sz(x) (int)x.size()
const ll maxn=2e5+5;
const ll maxlg=__lg(maxn)+2;
const ll INF64=4e18;
const int INF=0x3f3f3f3f;
const ll MOD=(1e9+7);
const ll MOD2=1000002173;
const ld PI=acos(-1);
const ld eps=1e-9;
#define lowb(x) x&(-x)
#define MNTO(x,y) x=min(x,(__typeof__(x))y)
#define MXTO(x,y) x=max(x,(__typeof__(x))y)
#define SORT_UNIQUE(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define GET_POS(c,x) (lower_bound(c.begin(),c.end(),x)-c.begin())
ll mult(ll a,ll b){
    return ((a%MOD)*(b%MOD))%MOD;
}
ll mypow(ll a,ll b,ll MOD){
    if(b<=0) return 1;
    ll res=1LL;
    while(b){
        if(b&1) res=(res*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return res;
}
pii a[maxn],b[maxn];
int ans[maxn];
int main(){
    ios::sync_with_stdio(false),cin.tie(0);
    int n;
    cin>>n;
    REP(i,n+1) cin>>a[i].f,a[i].s=i;
    REP(i,n) cin>>b[i].f,b[i].s=i;
    sort(a,a+n+1);
    sort(b,b+n);
    multiset<int> ms;
    REP(i,n){
        ms.insert(max(a[i+1].f-b[i].f,0));
    }
    ans[a[0].s]=(*prev(ms.end()));
    REP1(i,n){
        ms.erase(ms.find(max(a[i].f-b[i-1].f,0)));
        ms.insert(max(a[i-1].f-b[i-1].f,0));
        ans[a[i].s]=(*prev(ms.end()));
    }
    REP(i,n+1) cout<<ans[i]<<' ';
}
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