제출 #59775

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
59775		Benq	Swap (BOI16_swap)	C++11	68 / 100	825 ms	61696 KiB

swap

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

map<int,vi> bes[MX];
int n, p[MX];

vi merge(vi a, vi b) {
    assert(sz(a) >= sz(b));
    vi res;
    for (int x = 1; x <= sz(a); x *= 2) {
        FOR(i,x-1,min(2*x-1,sz(a))) res.pb(a[i]);
        FOR(i,x-1,min(2*x-1,sz(b))) res.pb(b[i]);
    }
    assert(sz(res) == sz(a)+sz(b));
    return res;
}

vi solve(int ind, int val) {
    if (bes[ind].count(val)) return bes[ind][val];
    if (2*ind > n) return {val};
    if (2*ind+1 > n) return {min(p[2*ind],val),max(p[2*ind],val)};
    
    vi v;
    if (val < min(p[2*ind],p[2*ind+1])) {
        v = merge(solve(2*ind,p[2*ind]),solve(2*ind+1,p[2*ind+1]));
        v.insert(v.begin(),val);
    } else if (p[2*ind] < min(val,p[2*ind+1])) {
        v = merge(solve(2*ind,val),solve(2*ind+1,p[2*ind+1]));
        v.insert(v.begin(),p[2*ind]);
    } else {
        v = min(merge(solve(2*ind,val),solve(2*ind+1,p[2*ind])),
                merge(solve(2*ind,p[2*ind]),solve(2*ind+1,val)));
        v.insert(v.begin(),p[2*ind+1]);
    }
    
    return bes[ind][val] = v;
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n;
    FOR(i,1,n+1) cin >> p[i];
    vi v = solve(1,p[1]);
    for (int i: v) cout << i << " ";
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 10 / 10
• Subtask 2: 11 / 11
• Subtask 3: 27 / 27
• Subtask 4: 20 / 20
• Subtask 5: 0 / 32