제출 #653054

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
653054		ayalla	여행하는 상인 (APIO17_merchant)	C++14	100 / 100	110 ms	2256 KiB

merchant

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define int long long int
#define endl '\n'
#define pb push_back
#define pi pair<int, int>
#define pii pair<int, pi>
#define fir first
#define sec second
#define MAXN 100005
#define mod 1000000007

const int inf = LLONG_MAX / 2;

int n, m, k;
int b[101][1001];
int s[101][1001];
int cost[101][101];
int stonks[101][101];
int g[101][101];
vector<int> adj[101];

void floyd(int dist[101][101])
{
  for (int k = 0; k < n; k++)
  {
    for (int i = 0; i < n; i++)
    {
      for (int j = 0; j < n; j++)
        dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
    }
  }
}
bool can(int mid)
{
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; j < n; j++)
      g[i][j] = -(stonks[i][j] - (min(cost[i][j], inf / mid) * mid));
  }
  // nesse grafo, agr quero saber se existe um ciclo que a soma dos custos das arestas eh >= 0
  floyd(g);
  for (int i = 0; i < n; i++)
  {
    if (g[i][i] <= 0)
      return 1;
  }
  return 0;
}
signed main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  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 < n; i++)
  {
    for (int j = 0; j < n; j++)
      cost[i][j] = inf;
  }
  for (int j = 0; j < m; j++)
  {
    int a, b, c;
    cin >> a >> b >> c;
    a--, b--;
    adj[a].pb(b);
    cost[a][b] = c;
  }
  floyd(cost);
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; j < n; j++)
    {
      for (int x = 0; x < k; x++)
      {
        if (s[j][x] == -1 || b[i][x] == -1)
          continue;
        stonks[i][j] = max(stonks[i][j], s[j][x] - b[i][x]);
      }
    }
  }
  int l = 0, r = 1e18;
  while (l < r)
  {
    int mid = (l + r + 1) >> 1;
    (can(mid)) ? l = mid : r = mid - 1;
  }
  cout << l << endl;
  return 0;
}
// n mercados
// m arestas de mao unica
// k diferentes items
// cada item tem
// b[i][j] - buy o item j no mercado i
// s[i][j] - sell o item j no mercado i
// arestas de v pra w usando t de tempo

// uma ideia eh
// pensar em resolver pra cada componente fortemente conexa

// busca binaria na resposta
// p / t >= mid
// entao
// p - (t * mid) >= 0

// 0 3 2 0

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