제출 #211886

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
211886		mode149256	Poi (IOI09_poi)	C++14	100 / 100	303 ms	16248 KiB

poi

/*input
5 3 2
0 0 1
1 1 0
1 0 0
1 1 0
1 1 0
*/
#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 {
	int sum;
	int kiek;
	int id;
};

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int N, T, P;
	cin >> N >> T >> P;

	vii sk(N, vi(T));
	vi neis(T, 0);

	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < T; ++j)
		{
			cin >> sk[i][j];
			neis[j] += !sk[i][j];
		}
	}

	vector<ppl> people(N);

	for (int i = 0; i < N; ++i)
	{
		auto &p = people[i];
		p.id = i + 1;
		p.sum = p.kiek = 0;

		for (int j = 0; j < T; ++j)
		{
			if (sk[i][j]) {
				p.kiek++;
				p.sum += neis[j];
			}
		}
	}

	sort(people.begin(), people.end(), [](const ppl a, const ppl b) {
		return a.sum > b.sum or (a.sum == b.sum and a.kiek > b.kiek) or
		       (a.sum == b.sum and a.kiek == b.kiek and a.id < b.id);
	});

	for (int i = 0; i < N; ++i)
	{
		if (people[i].id == P) {
			printf("%d %d\n", people[i].sum, i + 1);
			break;
		}
	}
}

/* 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: 100 / 100