제출 #496904

#제출 시각아이디문제언어결과실행 시간메모리
496904armashkaCoin Collecting (JOI19_ho_t4)C++17
0 / 100
1 ms332 KiB
#include <bits/stdc++.h> //#pragma comment(linker, "/stack:200000000") //#pragma GCC optimize("Ofast") //#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native") #define fast ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr); #define all(v) v.begin(),v.end() #define pb push_back #define sz size() #define ft first #define sd second using namespace std; typedef long long ll; typedef pair<ll, ll> pll; typedef unsigned long long ull; const int N = 1e6 + 5; const ll M = 1e8; const ll inf = 1e18; const ll mod = 1e9; const double Pi = acos(-1); ll binpow(ll x, ll ti) { ll res = 1;while (ti){if(ti & 1)res *= x;x *= x;ti >>= 1; x %= mod; res %= mod;} return res;} ll binmul(ll x, ll ti) { ll res = 0;while (ti){if(ti & 1)res += x;x += x;ti >>= 1; x %= mod; res %= mod;} return res;} ll nok(ll a, ll b) { return (a*b)/__gcd(abs(a),abs(b)) * (a*b > 0 ? 1 : -1); } bool odd(ll n) { return (n % 2 == 1); } bool even(ll n) { return (n % 2 == 0); } ll n, x[N], y[N], cnt[N][3], ans; vector <pll> v; const void solve(/*Armashka*/) { cin >> n; //cout << "\n"; for (int i = 1; i <= 2 * n; ++ i) { cin >> x[i] >> y[i]; if (y[i] > 2) ans += abs(y[i] - 2), y[i] = 2; if (y[i] < 1) ans += abs(y[i] - 1), y[i] = 1; if (x[i] > n) ans += abs(x[i] - n), x[i] = n; if (x[i] < 1) ans += abs(x[i] - 1), x[i] = 1; ++ cnt[x[i]][y[i]]; //cout << x[i] << " " << y[i] << "\n"; v.pb({x[i], y[i]}); } sort(all(v)); //cout << ans << "\n"; for (int j = 1; j <= 2; ++ j) { for (int i = 1; i <= n; ++ i) { if (cnt[i][j]) continue; ll cur = -1, mn = inf; for (int k = 0; k < 2 * n; ++ k) { ll res = abs(v[k].ft - i) + abs(v[k].sd - j); if (cnt[v[k].ft][v[k].sd] > 1 && res < mn) mn = res, cur = k; else if (cur != -1 && res == mn && j == v[k].sd && j != v[cur].sd) cur = k; } //cout << i << " " << j << " = " << v[cur].ft << " " << v[cur].sd << "\n"; ans += mn; -- cnt[v[cur].ft][v[cur].sd]; } } cout << ans << "\n"; } signed main() { fast; //freopen("divide.in", "r", stdin); //freopen("divide.out", "w", stdout); int tt = 1; //cin >> tt; while (tt --) { solve(); } } /* 5 4 4 1 2 3 1 3 4 5 3 4 5 2 3 2 1 3 1 1 3 5 2 3 4 5 2 1 3 1 */
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