Submission #127190

#	Time	Username	Problem	Language	Result	Execution time	Memory
127190		RockyB	Coin Collecting (JOI19_ho_t4)	C++17	0 / 100	276 ms	2040 KiB

joi2019_ho_t4

/// In The Name Of God

//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
//#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>

#define f first
#define s second

#define pb push_back
#define pp pop_back
#define mp make_pair

#define sz(x) (int)x.size()
#define sqr(x) ((x) * 1ll * (x))
#define all(x) x.begin(), x.end()

#define rep(i, l, r) for (int i = (l); i <= (r); i++)
#define per(i, l, r) for (int i = (l); i >= (r); i--)

#define Kazakhstan ios_base :: sync_with_stdio(0), cin.tie(0), cout.tie(0);

#define nl '\n'
#define ioi exit(0);

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

const int N = (int)2e5 + 7;
const int inf = (int)1e9 + 7;
const int mod = (int)1e9 + 7;
const ll linf = (ll)1e18 + 7;

const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};

using namespace std;

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

ll dp[N];
ll calc(int v = 1) {
  if (v > n) return 0;
  if (~dp[v]) return dp[v];

  ll res = linf;

  vector < pair <int, int > > put;
  vector <int> b;
  int ptr = 2 * v - 1;
  rep(i, 0, 4) {
    if (v + i > n) break;
    put.pb({v + i, 1});
    put.pb({v + i, 2});
    b.pb(ptr + 2 * i);
    b.pb(ptr + 2 * i + 1);
  }

  int f = 1;
  rep(i, 1, sz(b)) f *= i;

  rep(t, 1, f) {
    ll cur = 0;
    rep(i, 0, sz(b) - 1) {
      pair <ll, ll> x = a[b[i]];
      pair <ll, ll> y = put[i];
      cur += abs(x.f - y.f) + abs(x.s - y.s);
    }
    res = min(res, cur);
    next_permutation(all(put));
  }

  return dp[v] = res + calc(v + 5);
}
int main() {
	#ifdef IOI
		freopen ("in.txt", "r", stdin);
	#endif
  Kazakhstan
  cin >> n;
  rep(i, 1, n << 1) {
    cin >> a[i].f >> a[i].s;
  }
  sort (a + 1, a + n + n + 1);
  memset(dp, -1, sizeof(dp));
  cout << calc();
	ioi
}

• Subtask 1: 0 / 8
• Subtask 2: 0 / 29
• Subtask 3: 0 / 63