제출 #212425

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
212425		mode149256	Hiring (IOI09_hiring)	C++14	100 / 100	622 ms	48040 KiB

hiring

/*input
17 5862
19418 11899
688 748
846 909
506 684
516 368
265 802
464 55
14014 9205
6089 4279
613 481
13911 10444
11073 9160
141 643
4205 2955
13153 11233
971 16
7553 5487

*/
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}

struct ppl {
	ll S, Q;
	int i;
};

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int N;
	ll W;
	cin >> N >> W;
	vector<ppl> sk(N);

	for (int i = 0; i < N; ++i)
	{
		int a, b; cin >> a >> b;
		sk[i] = {a, b, i + 1};
	}

	sort(sk.rbegin(), sk.rend(), [&](const ppl & a, const ppl & b) {
		return a.Q * b.S < b.Q * a.S;
	});

	int kiek = 0;
	int l = 1;
	ll sumokejau = MOD;

	multiset<ll> buvo;
	ll sumQ = 0;

	if (sk[0].Q <= W) {
		kiek = 1;
		l = 0;
		sumokejau = sk[0].Q;
	}

	buvo.insert(sk[0].Q);

	// it first not included [0; it-1] it
	auto it = buvo.begin();
	int j = 0;

	for (int i = 1; i < N; ++i)
	{
		sumQ += sk[i].Q;
		while (j and sumQ * sk[i].S > W * sk[i].Q) {
			j--;
			it--;
			sumQ -= *it;
		}

		while (j + 1 <= i and (sumQ + *it) * sk[i].S <= W * sk[i].Q) {
			sumQ += *it;
			it++;
			j++;
		}

		if (sumQ * sk[i].S <= W * sk[i].Q) {
			if (j + 1 > kiek
			        or (j + 1 == kiek
			            and sumQ * sk[i].S * sk[l].Q < sumokejau * sk[i].Q)) {

				kiek = j + 1;
				l = i;
				sumokejau = sumQ * sk[i].S;
			}
		}

		buvo.insert(sk[i].Q);
		sumQ -= sk[i].Q;
		if (it == buvo.end() or sk[i].Q < *it) {
			sumQ += sk[i].Q;
			j++;
		}
	}

	vector<ppl> ans;
	for (int i = 0; i < l; ++i)
		ans.push_back(sk[i]);

	sort(ans.begin(), ans.end(), [&](const ppl a, const ppl b) {
		return a.Q < b.Q;
	});

	printf("%d\n", kiek);
	printf("%d\n", sk[l].i);
	for (int i = 0; i < kiek - 1; ++i)
		printf("%d\n", ans[i].i);
	// printf("l = %d, id = %d\n", l, sk[l].i);
}

/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/

• Subtask 1: 30 / 30
• Subtask 2: 28 / 28
• Subtask 3: 6 / 6
• Subtask 4: 6 / 6
• Subtask 5: 6 / 6
• Subtask 6: 6 / 6
• Subtask 7: 6 / 6
• Subtask 8: 6 / 6
• Subtask 9: 6 / 6