Submission #130011

#	Time	Username	Problem	Language	Result	Execution time	Memory
130011		qkxwsm	Naan (JOI19_naan)	C++14	100 / 100	610 ms	117116 KiB

naan

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

random_device(rd);
mt19937 rng(rd());
const long long FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();

struct custom_hash
{
	template<class T>
	unsigned long long operator()(T v) const
	{
		unsigned long long x = v;
		x += FIXED_RANDOM;
		// x += 11400714819323198485ull;
		// x = (x ^ (x >> 30)) * 13787848793156543929ull;
		x = (x ^ (x >> 27)) * 10723151780598845931ull;
		return x ^ (x >> 31);
	}
};

template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T, class U> using hash_table = gp_hash_table<T, U, custom_hash>;

template<class T>
T randomize(T mod)
{
	return (uniform_int_distribution<T>(0, mod - 1))(rng);
}
template<class T>
void readi(T &x)
{
	x = 0;
	bool negative = false;
	char c = ' ';
	while (c < '-')
	{
		c = getchar();
	}
	if (c == '-')
	{
		negative = true;
		c = getchar();
	}
	while (c >= '0')
	{
		x = x * 10 + (c - '0');
		c = getchar();
	}
	if (negative)
	{
		x = -x;
	}
}
template<class T>
void printi(T output)
{
	if (output == 0)
	{
		putchar('0');
		return;
	}
	if (output < 0)
	{
		putchar('-');
		output = -output;
	}
	int buf[20], n = 0;
	while(output)
	{
		buf[n] = ((output % 10));
		output /= 10;
		n++;
	}
	for (n--; n >= 0; n--)
	{
		putchar(buf[n] + '0');
	}
	return;
}
template<class T>
void ckmin(T &a, T b)
{
	a = min(a, b);
}
template<class T>
void ckmax(T &a, T b)
{
	a = max(a, b);
}
long long expo(long long a, long long e, long long mod)
{
	return ((e == 0) ? 1 : ((expo(a * a % mod, e >> 1, mod)) * ((e & 1) ? a : 1) % mod));
}
template<class T, class U>
void nmod(T &x, U mod)
{
	if (x >= mod) x -= mod;
}
template<class T>
T gcd(T a, T b)
{
	return (b ? gcd(b, a % b) : a);
}

#define y0 ___y0
#define y1 ___y1
#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define DBG(x) cerr << #x << " = " << (x) << endl
#define SZ(x) ((int) ((x).size()))
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define ALL(x) (x).begin(), (x).end()

const long double PI = 4.0 * atan(1.0);
const long double EPS = 1e-7;

#define MAGIC 347
#define SINF 10007
#define CO 1000007
#define INF 1000000007
#define BIG 1000000931
#define LARGE 1696969696967ll
#define GIANT 2564008813937411ll
#define LLINF 2696969696969696969ll
#define MAXN 2013

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ld, ld> pdd;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef vector<pdd> vpd;

int N, L;
ll num[MAXN][MAXN];
ld grid[MAXN][MAXN];
ld taken[MAXN];
bitset<MAXN> inside;
vpl ans;
vi ord;

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0);
	cout << fixed << setprecision(10);
	cerr << fixed << setprecision(10);
	// if (fopen("file.in", "r"))
	// {
	// 	freopen ("file.in", "r", stdin);
	// 	freopen ("file.out", "w", stdout);
	// }
	cin >> N >> L;
	FOR(i, 0, N)
	{
		ll tot = 0;
		FOR(j, 0, L)
		{
			cin >> num[i][j];
			// DBG(num[i][j]);
			tot += num[i][j];
		}
		FOR(j, 0, L)
		{
			grid[i][j] = 1.0 * num[i][j] * N / tot;
			// cerr << grid[i][j] << ' ';
		}
		// cerr << endl;
	}
	ans.PB({0, 1});
	//try to minimize the amount accomplished in the time it takes!
	FOR(i, 0, N - 1)
	{
		FOR(j, 0, N)
		{
			taken[j] = 0.0;
		}
		pll p = ans.back();
		int iter = p.fi / p.se;
		// DBG(iter);
		// cerr << p.fi << ' ' << p.se << endl;
		pair<ld, int> best = {INF, N};
		while(iter < L)
		{
			//see what happens if you go all the way up to iter + 1!
			ld x = ((p.fi / p.se == iter) ? (1.0 * ((iter + 1) * (p.se) - p.fi) / p.se) : 1.0);
			FOR(j, 0, N)
			{
				if (inside[j]) continue;
				taken[j] += x * grid[j][iter];
				// DBG(taken[j]);
				if (taken[j] > 1.0 - EPS)
				{
					ckmin(best, {1.0 + iter - (taken[j] - 1.0) / grid[j][iter], j});
				}
			}
			if (best.se != N) break;
			iter++;
		}
		int idx = best.se;
		ord.PB(idx);
		inside[idx] = true;
		pll res;
		res.se = 4 * num[idx][iter] * N;
		res.fi = (ll) (best.fi * res.se + 1.0 - EPS);
		ans.PB(res);
		//smth out of grid[i][iter].fi
		//x / grid[i][iter].fi >= best.fi
		//x >= best.fi * grid[i][iter].fi
		//ok now find the guy who can use the least!
	}
	FOR(i, 0, N) if (!inside[i]) ord.PB(i);
	ans.erase(ans.begin());
	for (pll p : ans)
	{
		ll g = gcd(p.fi, p.se);
		p.fi /= g; p.se /= g;
		cout << p.fi << ' ' << p.se << '\n';
	}
	FOR(i, 0, SZ(ord))
	{
		if (i) cout << ' ';
		cout << ord[i] + 1;
	}
	cout << '\n';
	// cerr << "time elapsed = " << (clock() / (CLOCKS_PER_SEC / 1000)) << " ms" << endl;
	return 0;
}
/* READ READ READ
* int overflow, maxn too small, special cases (n=1?, two distinct?), cin.tie() interactive
* reread the problem, try small cases
* note down possible sources of error as you go
* do smth instead of nothing
*/

• Subtask 1: 5 / 5
• Subtask 2: 24 / 24
• Subtask 3: 71 / 71