제출 #317503

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
317503		caoash	Swap (BOI16_swap)	C++14	48 / 100	1038 ms	11140 KiB

swap

/*
Realize it's a binary tree. Then realize each node can't have that many distinct values.

do dp[node][value], since transition is sum of subtree = sum of depth = nlogn in a binary tree it (shouldn't) MLE/TLE, but
it does! probably because i use map... will fix soon
*/

#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

#include <bits/stdc++.h> 
using namespace std;
 
using ll = long long;
 
using vi = vector<int>;
#define pb push_back
#define rsz resize
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
 
using pi = pair<int,int>;
#define f first
#define s second
#define mp make_pair
 
const int MX = 200005;
 
struct hash_pair { 
    template <class T1, class T2> 
    size_t operator()(const pair<T1, T2>& p) const
    { 
        auto hash1 = hash<T1>{}(p.first); 
        auto hash2 = hash<T2>{}(p.second); 
        return hash1 ^ hash2; 
    } 
}; 

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

int n; int num[MX]; gp_hash_table<pi, vi, hash_pair> dp; 
 
void merge(vi &best, vi a, int x, vi b) {
    vi ret;
    ret.pb(x);
    int p1 = 0, p2 = 0;
    for (int i = 0; i < 36; i++) {
        if (p1 < sz(a)) {
            for (int j = 0; j < (1 << i); j++) {
                if (p1 < sz(a)) {
                    ret.pb(a[p1++]);
                }
                else {
                    break;
                }
            } 
        }
        if (p2 < sz(b)) {
            for (int j = 0; j < (1 << i); j++) {
                if (p2 < sz(b)) {
                    ret.pb(b[p2++]);
                } 
                else {
                    break;
                }
            }
        }
    }
    if (best.empty()) best = ret;
    else best = min(best, ret);
}

vi solve(int v, int c) {
    if (!dp[mp(v, c)].empty()) {
        return dp[mp(v, c)];
    }
    int l = 2 * v + 1, r = 2 * v + 2;
    if (l >= n) {
        return dp[mp(v, c)] = {c};
    }
    if (r >= n) {
        vi best; 
        vi fst = solve(l, num[l]);
        best.pb(c);
        for (int x : fst) best.pb(x);
        vi sec = solve(l, c);
        vi sbest;
        sbest.pb(num[l]);
        for (int x : sec) sbest.pb(x);
        return dp[mp(v, c)] = min(best, sbest);
    }
    vi best;
    // no swaps
    merge(best, solve(l, num[l]), c, solve(r, num[r]));
 
    // swap left
    if (l < n) {
        merge(best, solve(l, c), num[l], solve(r, num[r]));
    }
 
    // swap right
    if (r < n) {
        merge(best, solve(l, num[l]), num[r], solve(r, c));
    }
 
    // swap left, then right
    if (r < n) {
        merge(best, solve(l, c), num[r], solve(r, num[l]));
    }
    return dp[mp(v, c)] = best;
}
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> num[i];
    }
    solve(0, num[0]); 
    vi ans = dp[mp(0, num[0])];
    for (int i = 0; i < n; i++) {
        cout << ans[i] << ' ';
    }
    cout << '\n';
}

컴파일 시 표준 에러 (stderr) 메시지

swap.cpp:9: warning: ignoring #pragma GCC optimization [-Wunknown-pragmas]
    9 | #pragma GCC optimization ("O3")
      | 
swap.cpp:10: warning: ignoring #pragma GCC optimization [-Wunknown-pragmas]
   10 | #pragma GCC optimization ("unroll-loops")
      |

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