Submission #1129043

#	Time	Username	Problem	Language	Result	Execution time	Memory
1129043		pcheloveks	Toll (BOI17_toll)	C++17	0 / 100	3094 ms	4128 KiB

toll

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("avx,avx2,tune=native")
#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <fstream>
#include <climits>
#include <algorithm>
#include <cstring> // Для memset

#define endl '\n'
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pii;

const LL DIM = 50007;
const LL INF = INT_MAX;

LL n, k, o, m, v1, v2, t;

class edge {
public:
    LL to, w;

    edge(LL to_, LL w_) {
        to = to_;
        w = w_;
    }
};

vector < edge > v[DIM];

class tran {
public:
    LL** mat;
    tran() {
        mat = new LL* [k];
        for (int i = 0; i < k; i++) {
            mat[i] = new LL[k];
            for (int j = 0; j < k; j++) {
                mat[i][j] = INF;
            }
        }
    }
    ~tran() {
        for (int i = 0; i < k; i++) {
            delete[] mat[i];
        }
        delete[] mat;
    }
};

tran solve(LL L, LL R) {
    tran ans;
    if (R - L == 1) {
        for (int i = 0; i < k; i++) {
            LL val = L * k + i;
            for (auto to : v[val]) {
                ans.mat[i][to.to - (R * k)] = to.w;
            }
        }
        return ans;
    }

    LL mid = (L + R) / 2;

    tran m1 = solve(L, mid);
    tran m2 = solve(mid, R);

    for (int l = 0; l < k; l++) {
        for (int r = 0; r < k; r++) {
            for (int m = 0; m < k; m++) {
                ans.mat[l][r] = min(ans.mat[l][r], m1.mat[l][m] + m2.mat[m][r]);
            }
        }
    }

    return ans;
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> k >> n >> m >> o;

    for (int i = 1; i <= m; i++) {
        cin >> v1 >> v2 >> t;
        v[v1].push_back({ v2, t });
    }

    vector<LL> out;

    for (int i = 1; i <= o; i++) {
        cin >> v1 >> v2;

        if (v1 / k >= v2 / k || m == 0) {
            out.push_back(-1);
        }
        else {
            tran res = solve(v1/ k, v2 / k);
            LL result = res.mat[v1 - (v1 / k) * k][v2 - (v2 / k) * k];

            if (result >= INF) cout << -1 << endl;
            else cout << result << endl;
        }
    }

    return 0;
}

• Subtask 1: 0 / 7
• Subtask 2: 0 / 10
• Subtask 3: 0 / 8
• Subtask 4: 0 / 31
• Subtask 5: 0 / 44