Submission #45092

#	Time	Username	Problem	Language	Result	Execution time	Memory
45092		qoo2p5	One-Way Streets (CEOI17_oneway)	C++17	0 / 100	3 ms	504 KiB

oneway

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;

const int INF = (int) 1e9 + 1e6 + 123;
const ll LINF = (ll) 1e18 + 1e9 + 123;

#define rep(i, s, t) for (auto i = (s); i < (t); ++(i))
#define per(i, s, t) for (auto i = (s); i >= (t); --(i))
#define sz(x) ((int)(x).size())
#define all(x) (x).begin(), (x).end()
#define mp make_pair
#define pb push_back

bool mini(auto &x, const auto &y) {
	if (y < x) {
		x = y;
		return 1;
	}
	return 0;
}

bool maxi(auto &x, const auto &y) {
	if (y > x) {
		x = y;
		return 1;
	}
	return 0;
}

void run();

mt19937 rnd;

int main(int argc, char *argv[]) {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	rnd.seed(atoi(argv[1]));
	run();
	return 0;
}

const int T = 100;
const int N = 10;
const int M = 15;
const int P = 5;

struct E {
	int a, b;
};

int n, m;
E e[T];
bool can[T][T];
bool can_left[T], can_right[T];
int p;
int x[T], y[T];

void floyd() {
	rep(k, 1, n + 1) rep(i, 1, n + 1) rep(j, 1, n + 1) can[i][j] |= can[i][k] && can[k][j];
}

bool test(int mask, int bit) {
	return mask & (1 << bit);
}

void run() {
	bool test_ok = 0;
	do {
		n = rnd() % N + 2;
		m = rnd() % M;
		p = rnd() % P;
	
		rep(i, 0, m) {
			int u, v;
			do {
				u = rnd() % n + 1;
				v = rnd() % n + 1;
			} while (u == v);
			e[i] = {u, v};
		}
		rep(i, 0, p) {
			int u, v;
			do {
				u = rnd() % n + 1;
				v = rnd() % n + 1;
			} while (u == v);
			x[i] = u;
			y[i] = v;
		}
		
		test_ok = 0;
	
		rep(mask, 0, 1 << m) {
			memset(can, 0, sizeof can);
			rep(i, 0, m) {
				int u = e[i].a;
				int v = e[i].b;
				if (test(mask, i)) {
					swap(u, v);
				}
				can[u][v] |= 1;
			}
			floyd();
			bool ok = 1;
			rep(i, 0, p) {
				ok &= can[x[i]][y[i]];
			}
			if (!ok) continue;
			test_ok = 1;
			break;
		}
	} while (!test_ok);
	
	cout << n << " " << m << "\n";
	rep(i, 0, m) {
		cout << e[i].a << " " << e[i].b << "\n";
	}
	cout << p << "\n";
	rep(i, 0, p) {
		cout << x[i] << " " << y[i] << "\n";
	}
}

• Subtask 1: 0 / 30
• Subtask 2: 0 / 30
• Subtask 3: 0 / 40