제출 #274887

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
274887		infinitepro	여행하는 상인 (APIO17_merchant)	C++17	100 / 100	137 ms	8312 KiB

merchant

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

#define int ll

typedef vector<int> vi;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<vi> vvi;
typedef vector<vii> vvii;

#define INF INT_MAX
#define MOD 1000000007
#define all(x) x.begin(), x.end()

bool calc(vvi &dist, vvi &profit, int &N, int R){
    vvi G(N, vi(N, -INF));
    for(int u = 0; u < N; u++){
        for(int v = 0; v < N; v++){
            if(u == v || dist[u][v] == INF) continue;
            G[u][v] = profit[u][v] - R*dist[u][v];
        }
    }

    for(int i = 0; i < N; i++)
        for(int u = 0; u < N; u++)
            for(int v = 0; v < N; v++)
                G[u][v] = max(G[u][v], G[u][i]+G[i][v]);
    
    for(int x = 0; x < N; x++)
        if(G[x][x] >= 0) return true;
    return false;
}

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int N, M, K; cin >> N >> M >> K;
    vector<vvi> C(N, vvi(K, vi(2))); // C[i][k] = {buy, sell}
    for(int x = 0; x < N; x++)
        for(int k = 0; k < K; k++)
            cin >> C[x][k][0] >> C[x][k][1];
    
    vvi dist(N, vi(N, INF)), profit(N, vi(N));
    while(M--){
        int u, v, c; cin >> u >> v >> c;
        u--, v--; dist[u][v] = c;
    }
    for(int x = 0; x < N; x++) dist[x][x] = 0;

    for(int i = 0; i < N; i++)
        for(int u = 0; u < N; u++)
            for(int v = 0; v < N; v++)
                dist[u][v] = min(dist[u][v], dist[u][i]+dist[i][v]);

    for(int u = 0; u < N; u++){
        for(int v = 0; v < N; v++){
            for(int k = 0; k < K; k++){
                if(C[u][k][0] != -1 && C[v][k][1] != -1)
                    profit[u][v] = max(profit[u][v], C[v][k][1]-C[u][k][0]);
            }
        }
    }

    int low = 0, high = 1e9;
    while(low < high){
        int mid = (low+high+1)/2;
        if(calc(dist, profit, N, mid)) low = mid;
        else high = mid-1;
    }

    cout << low << endl;

    return 0;
}

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