Submission #544848

#	Time	Username	Problem	Language	Result	Execution time	Memory
544848		valerikk	Travelling Merchant (APIO17_merchant)	C++17	100 / 100	98 ms	4432 KiB

merchant

#include <bits/stdc++.h>

using namespace std;

#define int long long 

const int N = 105;
const int K = 1010;
const int M = 9999;
const long long INF = 0x3f3f3f3f3f3f3f3f;

int n, m, k;
int b[N][K];
int s[N][K];
int v[M], w[M], t[M];

int mx[N][N];
long long d[N][N];
long long dd[N][N];

signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m >> k;
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < k; ++j) {
			cin >> b[i][j] >> s[i][j];
		}
	}
	for (int i = 0; i < m; ++i) {
		cin >> v[i] >> w[i] >> t[i];
		--v[i];
		--w[i];
	}

	memset(d, 0x3f, sizeof(d));
	for (int i = 0; i < n; ++i) {
		d[i][i] = 0;
	}
	for (int i = 0; i < m; ++i) {
		d[v[i]][w[i]] = min<long long>(d[v[i]][w[i]], t[i]);
	}
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			for (int f = 0; f < n; ++f) {
				if (d[j][i] < INF && d[i][f] < INF) {
					d[j][f] = min(d[j][f], d[j][i] + d[i][f]);
				}
			}
		}
	}
	memcpy(dd, d, sizeof(dd));

	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			mx[i][j] = 0;
			for (int f = 0; f < k; ++f) {
				if (b[i][f] != -1 && s[j][f] != -1) {
					mx[i][j] = max(mx[i][j], s[j][f] - b[i][f]);
				}
			}
		}
	}
	
	int l = 0, r = 1e9;
	while (r - l > 1) {
		int mid = (l + r) / 2;
		
		memcpy(d, dd, sizeof(d));
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < n; ++j) {
				if (i != j && d[i][j] < INF) {
					d[i][j] = d[i][j] * mid - mx[i][j];
					assert(d[i][j] < INF);
				}
			}
		}
		for (int i = 0; i < n; ++i) {
			d[i][i] = INF;
		}
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < n; ++j) {
				for (int f = 0; f < n; ++f) {
					if (d[j][i] < INF && d[i][f] < INF) {
						d[j][f] = min(d[j][f], d[j][i] + d[i][f]);
					}
				}
			}
		}

		bool ok = false;
		for (int i = 0; i < n; ++i) {
			ok |= d[i][i] <= 0;
		}
		if (ok) {
			l = mid;
		} else {
			r = mid;
		}
	}

	cout << l << endl;

	return 0;
}

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