Submission #1138507

#	Time	Username	Problem	Language	Result	Execution time	Memory
1138507		pan	Coin Collecting (JOI19_ho_t4)	C++20	100 / 100	59 ms	7492 KiB

joi2019_ho_t4

#include <bits/stdc++.h>
//#include "includeall.h"
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define endl '\n'
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define input(x) scanf("%lld", &x);
#define input2(x, y) scanf("%lld%lld", &x, &y);
#define input3(x, y, z) scanf("%lld%lld%lld", &x, &y, &z);
#define input4(x, y, z, a) scanf("%lld%lld%lld%lld", &x, &y, &z, &a);
#define print(x, y) printf("%lld%c", x, y);
#define show(x) cerr << #x << " is " << x << endl;
#define show2(x,y) cerr << #x << " is " << x << " " << #y << " is " << y << endl;
#define show3(x,y,z) cerr << #x << " is " << x << " " << #y << " is " << y << " " << #z << " is " << z << endl;
#define all(x) x.begin(), x.end()
#define discretize(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define FOR(i, x, n) for (ll i =x; i<=n; ++i) 
#define RFOR(i, x, n) for (ll i =x; i>=n; --i) 
#pragma GCC optimize("O3","unroll-loops")
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
//using namespace __gnu_pbds;
//#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
//#define ordered_multiset tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
typedef long long ll;
typedef long double ld;
typedef pair<ld, ll> pd;
typedef pair<string, ll> psl;
typedef pair<ll, ll> pi;
typedef pair<ll, pi> pii;
typedef pair<pi, pi> piii;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());




int main()
{
	cin.tie(nullptr);
	cout.tie(nullptr);
	ios::sync_with_stdio(false);
	ll n, ans = 0;
	cin >> n;
	vector<vector<ll> > xy(n + 5, vector<ll> (5, 0));
	for (ll i=0; i<2*n; ++i)
	{
		ll x, y;
		cin >> x >> y;
		if (x > n) ans += (x - n), x = n;
		if (x < 1) ans += 1 - x, x = 1;
		if (y > 2) ans += y-2, y = 2;
		if (y < 1) ans += 1 - y, y=1;
		xy[x][y]++;
	}
	ll ps1 = 0, ps2 = 0;
	for (ll i=1; i<=n; ++i)
	{
		ps1 += xy[i][1] - 1, ps2 += xy[i][2] - 1;
		ll take = max(0LL, min(-ps1, ps2)); ps1 += take, ps2 -= take;ans += take;
		take = max(0LL, min(ps1, -ps2)); ps1 -= take, ps2 += take; ans += take;
		ans += abs(ps1) + abs(ps2);

	}
	cout << ans << endl;
	return 0;
}

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