제출 #555141

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
555141		nutella	여행하는 상인 (APIO17_merchant)	C++17	0 / 100	1089 ms	2168 KiB

merchant

//#define _GLIBCXX_DEBUG

//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

#include <bits/stdc++.h>

using namespace std;

//#include <ext/pb_ds/assoc_container.hpp>
//
//using namespace __gnu_pbds;
//
//template<typename T>
//using ordered_set = tree<T, null_type, less < T>, rb_tree_tag, tree_order_statistics_node_update>;

template<typename T>
using normal_queue = priority_queue<T, vector<T>, greater<>>;

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

#define ll long long
#define trace(x) cout << #x << ": " << (x) << endl;
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define uniq(x) x.resize(unique(all(x)) - begin(x))
#define ld long double
#define sz(s) ((int) size(s))
#define pii pair<int, int>
#define mp(x, y) make_pair(x, y)
#define int128 __int128
#define pb push_back
#define eb emplace_back
#define fi first
#define se second


template<typename T>
bool ckmn(T &x, T y) {
    if (x > y) {
        x = y;
        return true;
    }
    return false;
}

template<typename T>
bool ckmx(T &x, T y) {
    if (x < y) {
        x = y;
        return true;
    }
    return false;
}

int bit(int x, int b) {
    return (x >> b) & 1;
}

int rand(int l, int r) { return (int) ((ll) rnd() % (r - l + 1)) + l; }

const ll infL = 3e18;
const int infI = 1000000000 + 7;
const int infM = 2139062143;
const ll infML = 9187201950435737471LL;
const int maxN = 101, maxK = 1001;
const ll mod = 998244353;
const ld eps = 1e-9;

int N, M, K, b[maxN][maxK], s[maxN][maxK], dist[maxN][maxN], bb[maxN][maxN];
pair<ll, ll> dp[2][maxN];

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    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];
        }
        fill(dist[i], dist[i] + N, infI);
        dist[i][i] = 0;
    }
    for (int i = 0; i < M; ++i) {
        int a, v, w;
        cin >> a >> v >> w;
        --a, --v;
        dist[a][v] = w;
    }
    for (int k = 0; k < N; ++k)
        for (int i = 0; i < N; ++i)
            for (int j = 0; j < N; ++j)
                if (dist[i][k] != infI && dist[k][j] != infI)
                    ckmn(dist[i][j], dist[i][k] + dist[k][j]);
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            for (int k = 0; k < K; ++k) {
                if (b[i][k] != -1 && s[j][k] != -1)
                    ckmx(bb[i][j], s[j][k] - b[i][k]);
            }
        }
    }
    ll ans = 0;
    for (int v = 0; v < N; ++v) {
        auto mxx = [](pair<ll, ll> &a, pair<ll, ll> b) {
            if (b == mp(-infL, -infL)) return;
            if (a == mp(-infL, -infL)) {
                a = b;
                return;
            }
            if (a.first != 0)
                assert(abs(b.second) <= abs(LONG_LONG_MAX / abs(a.first)));
            if (a.second != 0)
                assert(abs(b.first) <= abs(LONG_LONG_MAX / abs(a.second)));
            if (a.first * b.second < a.second * b.first) a = b;
        };
        for (int i = 0; i < N; ++i) {
            dp[0][i] = {-infL, -infL};
        }
        dp[0][v] = {0, 0};
        for (int mv = 1; mv < 2 * N + 3; ++mv) {
            int nw = mv & 1;
            int prv = nw ^ 1;
            for (int i = 0; i < N; ++i) {
                dp[nw][i] = {-infL, -infL};
                for (int j = 0; j < N; ++j) {
                    pair<ll, ll> now = dp[prv][j];
                    if (now == mp(-infL, -infL)) continue;
                    now.first += bb[j][i];
                    now.second += dist[j][i];
                    mxx(dp[nw][i], now);
                }
            }
            if (dp[nw][v] != mp(-infL, -infL) && dp[nw][v].first > 0)
                ckmx(ans, dp[nw][v].first / dp[nw][v].second);
        }
    }
    cout << ans;
    return 0;
}

• Subtask 1: 0 / 12
• Subtask 2: 0 / 21
• Subtask 3: 0 / 33
• Subtask 4: 0 / 34