Submission #756296

#	Time	Username	Problem	Language	Result	Execution time	Memory
756296		Shreyan_Paliwal	Toll (BOI17_toll)	C++17	7 / 100	179 ms	93316 KiB

toll

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

const int maxn = 50000;
const int maxo = 10000;
const int maxk = 5;
const int logn = 19;
const int INF = 1e9;

int k, n, m, o;
pair<int,int> orders[maxo];
int dp[maxn][logn][maxk][maxk]; // from k*maxn+maxk1 --> k*(maxn+2^logn) + maxk2

inline void comb(int a2[maxk][maxk], int a[maxk][maxk], int m[maxk][maxk]) {
    for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++)
            for (int l = 0; l < k; l++) 
                a2[i][l] = min(a2[i][l], a[i][j] + m[j][l]);
}

int main() {
    // freopen("main.in", "r", stdin);

    fill(dp[0][0][0], dp[0][0][0] + maxn*logn*maxk*maxk, INF);

    cin >> k >> n >> m >> o;
    for (int i = 0; i < m; i++) {
        int a, b, t; cin >> a >> b >> t;
        dp[a / k][0][a%k][b%k] = t;
    }

    // for (int i = 0; i <= 1; i++) {
    //     for (int a = 0; a < 5; a++) {
    //         for (int b = 0; b < 5; b++) {
    //             cout << dp[i][0][a][b] << ' ';
    //         }
    //         cout << endl;
    //     }
    // }

    // set up binary lifting

    for (int i = 1; i < logn; i++) { // for each power of 2 jump
        for (int j = 0; j + (1 << i) <= (n-k+1)/k; j++) { // do all j where j+(power of 2) is still an actual section
            comb(dp[j][i], dp[j][i-1], dp[j+(1<<(i-1))][i-1]);
        }
    }

    int ans[maxk][maxk], ans2[maxk][maxk];
    for (int i = 0; i < o; i++) {
        int a, b; cin >> a >> b;
        int A = a/k, B = b/k;
        fill(ans[0], ans[0] + maxk*maxk, INF);
        for (int j = 0; j < k; j++) ans[j][j] = 0;
        // ans[a%k][a%k] = 0; // because we are starting at a%k, and only care about paths from it
        // a->b where a =/= b doesn't make sense since we haven't made jumps, and rest we aren't starting at it
        for (int j = logn-1; j >= 0; j--)
            if (((B-A)>>j) & 1) {
                // make jump of size j
                fill(ans2[0], ans2[0] + maxk*maxk, INF);
                comb(ans2, ans, dp[A][j]);
                A += (1<<j);
                for (int c = 0; c < k; c++)
                    for (int d = 0; d < k; d++)
                        ans[c][d] = ans2[c][d];
            }
        cout << (ans[a%k][b%k] == INF ? -1 : ans[a%k][b%k]) << endl;
    }

    return 0;
}

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