제출 #556672

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
556672		hollwo_pelw	여행하는 상인 (APIO17_merchant)	C++17	100 / 100	96 ms	3340 KiB

merchant

#include <bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/trie_policy.hpp>
// #include <ext/rope>

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

void Hollwo_Pelw();

signed main(){
#ifndef hollwo_pelw_local
	if (fopen("hollwo_pelw.inp", "r"))
		assert(freopen("hollwo_pelw.inp", "r", stdin)), assert(freopen("hollwo_pelw.out", "w", stdout));
#else
	using namespace chrono;
	auto start = steady_clock::now();
#endif
	cin.tie(0), cout.tie(0) -> sync_with_stdio(0);
	int testcases = 1;
	// cin >> testcases;
	for (int test = 1; test <= testcases; test++){
		// cout << "Case #" << test << ": ";
		Hollwo_Pelw();
	}
#ifdef hollwo_pelw_local
	auto end = steady_clock::now();
	cout << "\nExcution time : " << duration_cast<milliseconds> (end - start).count() << "[ms]" << endl;
#endif
}

const int N = 105, K = 1005, inf = 1e9 + 7;
const long long linf = 4e18;

int n, m, k, b[N][K], s[N][K], d[N][N], c[N][N];
long long f[N][N];

void Hollwo_Pelw(){
	cin >> n >> m >> k;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= k; j++)
			cin >> b[i][j] >> s[i][j];

	for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++)
		for (int _ = 1; _ <= k; _++) if (s[j][_] >= 0 && b[i][_] >= 0)
			c[i][j] = max(c[i][j], s[j][_] - b[i][_]);

	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++)
			d[i][j] = i == j ? 0 : inf;

	for (int u, v, w; m --; )
		cin >> u >> v >> w, d[u][v] = w;

	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				d[i][j] = min(d[i][j], d[i][k] + d[k][j]);

	int l = 0, r = 1e9 + 7;

	while (l < r) {
		int m = (l + r + 1) >> 1, ok = 0;
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++) {
				f[i][j] = d[i][j] == inf || i == j ? linf : 1ll * d[i][j] * m - c[i][j];
			}
		for (int k = 1; k <= n; k++)
			for (int i = 1; i <= n; i++)
				for (int j = 1; j <= n; j++)
					f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
		for (int i = 1; i <= n && !ok; i++)
			if (f[i][i] <= 0) ok = 1;

		(ok ? l = m : r = m - 1);
	}

	cout << l << '\n';
}

• Subtask 1: 12 / 12
• Subtask 2: 21 / 21
• Subtask 3: 33 / 33
• Subtask 4: 34 / 34