Submission #540823

#	Time	Username	Problem	Language	Result	Execution time	Memory
540823		Lobo	Travelling Merchant (APIO17_merchant)	C++17	100 / 100	73 ms	4424 KiB

merchant

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

const long long inf = (long long) 1e18 + 10;
const int inf1 = (int) 1e9 + 10;
#define int long long
#define dbl long double
#define endl '\n'
#define sc second
#define fr first
#define mp make_pair
#define pb push_back
#define all(x) x.begin(), x.end()

const int maxn = 110;
const int maxk = 1100;

int n,m,k, b[maxn][maxk], s[maxn][maxk], dfw[maxn][maxn], d[maxn][maxn], g[maxn][maxn], a[maxn][maxn];

void solve() {
    cin >> n >> m >> k;

    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= k; j++) {
            cin >> b[i][j];
            if(b[i][j] == -1) b[i][j] = inf;
            cin >> s[i][j];
            if(s[i][j] == -1) s[i][j] = -inf;
        }
    }

    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= n; j++) {
            dfw[i][j] = inf1;
        }
    }
    for(int i = 1; i <= m; i++) {
        int u,v,w; cin >> u >> v >> w;
        dfw[u][v] = min(dfw[u][v],w);
    }

    for(int v = 1; v <= n; v++) {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                dfw[i][j] = min(dfw[i][j], dfw[i][v]+dfw[v][j]);
            }
        }
    }

    for(int u = 1; u <= n; u++) {
        for(int v = 1; v <= n; v++) {
            a[u][v] = 0;
            //compra em u e vende em v
            for(int i = 1; i <= k; i++) {
                a[u][v] = max(a[u][v],s[v][i]-b[u][i]);
            }
            // cout << u << " " << v << " " << dfw[u][v] << endl;
        }
    }

    int l = 1;
    int r = inf1; //valor maximo
    int ans = 0;
    while(l <= r) {
        int mid = (l+r)>>1;

        //cria as arestas
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                d[i][j] = -inf1;
            }
        }

        for(int u = 1; u <= n; u++) {
            for(int v = 1; v <= n; v++) {
                g[u][v] = a[u][v]-dfw[u][v]*mid;
                d[u][v] = g[u][v];
                // cout << u << " " << v << " " << d[u][v] << endl;
            }
        }

        bool ok = false;

        for(int v = 1; v <= n; v++) {
            for(int i = 1; i <= n; i++) {
                for(int j = 1; j <= n; j++) {
                    d[i][j] = max(d[i][j], d[i][v]+d[v][j]);
                }
            }
        }

        for(int i = 1; i <= n; i++) {
            // cout << d[i][i] << endl;
            if(d[i][i] >= 0) ok = true;
        }

        //encontrar se tem ciclo positivo em um grafo bellman

        // bool ok = false;
        // for(int beg = 1; beg <= n; beg++) {
        //     for(int i = 1; i <= n; i++) {
        //         d[i][0] = -inf;
        //     }
        //     d[beg][0] = 0;
        //     for(int i = 1; i <= n+1; i++) {
        //         for(int u = 1; u <= n; u++) d[u][i] = d[u][i-1];

        //         for(int u = 1; u <= n; u++) {
        //             for(int v = 1; v <= n; v++) {
        //                 d[v][i] = max(d[v][i],d[u][i-1] + g[u][v]);
        //             }
        //         }
        //     }

        //     for(int u = 1; u <= n; u++) {
        //         for(int v = 1; v <= n; v++) {
        //             if(d[u][n] != -inf && d[v][n] < d[u][n]+g[u][v]) {
        //                 cout << beg << " " << u << " " << v << " " << d[u][n] << " " << d[v][n] << " " << g[u][v] << endl;
        //                 ok = true;
        //             }
        //             // d[v][i] = max(d[v][i],d[u][i-1] + g[u][v]);
        //         }
        //     }
        // }

        if(ok) {
            ans = mid;
            l = mid+1;
        }
        else {
            r = mid-1;
        }
    }

    cout << ans << endl;
}

int32_t main() {
    ios::sync_with_stdio(false); cin.tie(0);

    // freopen("in.in", "r", stdin);
    // freopen("out.out", "w", stdout);

    int tt = 1;
    // cin >> tt;
    while(tt--) solve();

}

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