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in cpp with short variables

Réflexion

Understanding the coding problem details

The user provided images of a coding problem about visiting cities with red and blue lines, where switching to or starting from the blue line incurs a "blueCost." It seems we are working with n+1 cities where n is the size of the red and blue arrays, depicting travel times between each city. I'll need to implement this using C++.

Planning the dynamic programming approach

We need to find the minimum time from city 0 to each city, returning an array that holds the times for each city. For this, I’ll use dynamic programming with two states: one for being on the red line and another for the blue line. For transitions, there’s a cost to switch to blue, but no extra cost for using red.

Finalizing the dynamic programming solution

If we start on the blue line, we pay the blueCost once. We set dpB[0] = blueCost and dpR[0] = 0. For each city, we update the times on the red or blue lines using dynamic programming. I'll output the results for each city, ensuring efficiency given the constraints. I'll provide the C++ solution skeleton next.

Confirming the DP solution and preparing C++ code

The dynamic programming approach works. For each test case, I’ve checked the sample outputs and verified the calculations. Everything is correct, so now I'll implement the solution. For the C++ function, I’ll make sure to follow the request for short variable names and return the correct vector length. Time to write the code.

Thought for 54s

Réponse

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

vector<long long> minimumCost(vector<int> r, vector<int> b, int c) {
    int n = (int)r.size();
    vector<long long> a(n + 1);
    long long rr = 0, bb = (long long)c;   // cost at city 0 on red / blue
    a[0] = 0;

    for (int i = 1; i <= n; i++) {
        long long nr = min(rr, bb) + (long long)r[i - 1];                       // switch to red is free
        long long nb = min(bb + (long long)b[i - 1], rr + (long long)c + (long long)b[i - 1]); // pay c if coming from red
        rr = nr; bb = nb;
        a[i] = min(rr, bb);
    }
    return a;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n; 
    cin >> n;
    vector<int> r(n);
    for (int i = 0; i < n; i++) cin >> r[i];

    int m; 
    cin >> m;
    vector<int> b(m);
    for (int i = 0; i < m; i++) cin >> b[i];

    int c; 
    cin >> c;

    auto a = minimumCost(r, b, c);
    for (auto x : a) cout << x << "\n";
    return 0;
}

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