제출 #470993

#제출 시각아이디문제언어결과실행 시간메모리
470993NamnamseoNaan (JOI19_naan)C++17
29 / 100
4018 ms82356 KiB
#include <algorithm>
#include <iostream>
#include <numeric>
using namespace std;
using ll=long long;
const int maxn = int(2e3) + 10;

int n, l;
ll v[maxn][maxn];
ll vs[maxn];

ll gcd(ll a, ll b) { while (b) a %= b, swap(a, b); return a; }

struct frac {
	ll m, j;
	frac(ll j_=0, ll m_=1) : m(m_), j(j_) {
		if (m < 0) m=-m, j=-j;
		ll g = gcd(m, abs(j));
		m /= g; j /= g;
	}
	explicit operator int() { return j/m; }
	explicit operator ll() { return j/m; }
	bool operator<(frac r) const { return j*r.m < m*r.j; }
	bool operator>(frac r) const { return j*r.m > m*r.j; }
	bool operator<=(frac r) const { return j*r.m <= m*r.j; }
	bool operator>=(frac r) const { return j*r.m >= m*r.j; }
	frac operator+(frac r) { return frac(j*r.m + m*r.j, m*r.m); }
	frac operator-(frac r) { return frac(j*r.m - m*r.j, m*r.m); }
	frac operator*(frac r) { return frac(j*r.j, m*r.m); }
	frac operator/(frac r) { return frac(j*r.m, m*r.j); }
	frac ceil_to(ll nm) { return frac((j*nm+m-1)/m, nm); }
};

int ansp[maxn];
bool done[maxn];

frac cuts[maxn][maxn];

int main() { cin.tie(0)->sync_with_stdio(0);
	cin >> n >> l;
	for (int i=1; i<=n; ++i) for (int j=1; j<=l; ++j) cin >> v[i][j];
	for (int i=1; i<=n; ++i) vs[i] = accumulate(v[i]+1, v[i]+l+1, 0ll);

	for (int i=1; i<=n; ++i) {
		frac cx{};
		frac tgt{vs[i], n};
		int cj = 0;
		for (int j=1; j<=l; ++j) {
			frac yb{v[i][j]};
			while (yb.j) {
				if (tgt <= yb) {
					yb = yb - tgt;
					cx = cx + tgt / frac{v[i][j]};
					if (cx.m >= int(1e9)) for(;;);
					cuts[i][++cj] = cx;
					tgt = {vs[i], n};
				} else {
					tgt = tgt - yb;
					cx = j;
					yb = 0;
				}
			}
		}
	}

	frac cx{};
	for (int pi=1; pi<=n; ++pi) {
		frac mx; int mxi = -1;
		for (int i=1; i<=n; ++i) if (!done[i]) {
			frac tx = cuts[i][pi];
			if (mxi == -1 || tx < mx) mx = tx, mxi = i;
		}
		cx = mx;
		if (pi < n) cout << mx.j << ' ' << mx.m << '\n';
		ansp[pi] = mxi;
		done[mxi] = true;
	}

	for (int i=1; i<=n; ++i) cout << ansp[i] << " \n"[i == n];

	return 0;
}
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