Submission #672664

#	Time	Username	Problem	Language	Result	Execution time	Memory
672664		ghostwriter	Coin Collecting (JOI19_ho_t4)	C++17	0 / 100	8 ms	380 KiB

joi2019_ho_t4

// #pragma GCC optimize ("Ofast")
// #pragma GCC target ("avx2")
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
    #include <debug.h>
#else
    #define debug(...)
#endif
#define ft front
#define bk back
#define st first
#define nd second
#define ins insert
#define ers erase
#define pb push_back
#define pf push_front
#define _pb pop_back
#define _pf pop_front
#define lb lower_bound
#define ub upper_bound
#define mtp make_tuple
#define bg begin
#define ed end
#define all(x) (x).bg(), (x).ed()
#define sz(x) (int)(x).size()
// #define int long long
typedef long long ll; typedef unsigned long long ull;
typedef double db; typedef long double ldb;
typedef pair<int, int> pi; typedef pair<ll, ll> pll;
typedef vector<int> vi; typedef vector<ll> vll; typedef vector<pi> vpi; typedef vector<pll> vpll;
typedef string str;
#define FOR(i, l, r) for (int i = (l); i <= (r); ++i)
#define FOS(i, r, l) for (int i = (r); i >= (l); --i)
#define FRN(i, n) for (int i = 0; i < (n); ++i)
#define FSN(i, n) for (int i = (n) - 1; i >= 0; --i)
#define EACH(i, x) for (auto &i : (x))
#define WHILE while
template<typename T> T gcd(T a, T b) { T d2 = (a | b) & -(a | b); a /= d2; b /= d2; WHILE(b) { a = a % b; swap(a, b); } return a * d2; }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
void _assert(bool statement) { if (statement) return; cerr << "\n>> Assertion failed!\n"; exit(0); }
void _assert(bool statement, const str &message) { if (statement) return; cerr << "\n>> Assertion failed: " << message << '\n'; exit(0); }
void _error(const str &message) { cerr << "\n>> Error: " << message << '\n'; exit(0); }
#define file "TEST"
mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
ll rand(ll l, ll r) { return uniform_int_distribution<ll>(l, r)(rd); }
/*
----------------------------------------------------------------
    END OF TEMPLATE
----------------------------------------------------------------
    Tran The Bao - ghostwriter
    Training for VOI23 gold medal
----------------------------------------------------------------
    GOAT
----------------------------------------------------------------
*/
const int N = 2e5 + 5;
int n;
pi a[N];
void input(int test_id) {
	cin >> n;
	FOR(i, 1, n << 1) cin >> a[i].st >> a[i].nd;
}
ll cost(const pi &a, const pi &b) { return 1LL * abs(a.st - b.st) + abs(a.nd - b.nd); }
namespace subtask12 {
    const ll oo = 1e18;
    const int N = 2005;
    ll d[N][N];
    void solve() {
        n <<= 1;
        sort(a + 1, a + 1 + n);
        FOS(i, n, 1)
        FOS(j, n >> 1, 0) {
            d[i][j] = oo;
            if (j < n / 2) d[i][j] = min(d[i][j], d[i + 1][j + 1] + cost(a[i], {j + 1, 1}));
            if (i - 1 - j < n / 2) d[i][j] = min(d[i][j], d[i + 1][j] + cost(a[i], {i - j, 2}));
        }
        cout << d[1][0];
    }
}
namespace subtask3 {
    /*
        Interesting problem :v. The main idea is to match every coin with a grid coordinate. We can transform every coin that lie outside the box
        to a coin that lie inside the box and create a new problem that is so much easier!
    */
    int d[N][3];
    void solve() {
        ll rs = 0;
        FOR(i, 1, n << 1) {
            if (a[i].st < 1) {
                rs += 1 - a[i].st;
                a[i].st = 1;
            }
            if (a[i].st > n) {
                rs += a[i].st - n;
                a[i].st = n;
            }
            if (a[i].nd < 1) {
                rs += 1 - a[i].nd;
                a[i].nd = 1;
            }
            if (a[i].nd > 2) {
                rs += a[i].nd - 2;
                a[i].nd = 2;
            }
            ++d[a[i].st][a[i].nd];
        }
        int pos = 1;
        auto update = [&]() {
            WHILE(!d[pos][1] && !d[pos][2]) ++pos;
        };
        FOR(i, 1, n) {
            update();
            int npos = 1;
            if (d[pos][2]) npos = 2;
            rs += pos - i;
            --d[pos][npos];
            npos = 3 - npos;
            update();
            if (d[pos][npos]) {
                rs += pos - i;
                --d[pos][npos];
            }
            else {
                rs += pos - i + 1;
                --d[pos][3 - npos];
            }
            update();
            if (pos == i) {
                rs += d[pos][1] + d[pos][2];
                d[pos + 1][1] += d[pos][1];
                d[pos + 1][2] += d[pos][2];
                d[pos][1] = d[pos][2] = 0;
            }
        }
        cout << rs;
    }
}
void solve(int test_id) {
    // if (n <= 1000) {
    //     subtask12::solve();
    //     return;
    // }
    subtask3::solve();
}
void reinit(int test_id) {

}
signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    // freopen(file".inp", "r", stdin);
    // freopen(file".out", "w", stdout);
    int test_num = 1;
    // cin >> test_num; // comment if the problem does not requires multitest
    FOR(i, 1, test_num) {
        input(i); // input in noninteractive problems for case #i
        solve(i); // main function to solve case #i
        reinit(i); // reinit global data to default used in case #i
    }
    #ifdef LOCAL
        cerr << "\nTime: " << setprecision(5) << fixed << (ldb)clock() / CLOCKS_PER_SEC << "ms.\n";
    #endif
    return 0;
}
/*
----------------------------------------------------------------
From Benq:
    stuff you should look for
        * int overflow, array bounds
        * special cases (n=1?)
        * do smth instead of nothing and stay organized
        * WRITE STUFF DOWN
        * DON'T GET STUCK ON ONE APPROACH
----------------------------------------------------------------
*/

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