In this problem, we calculate the minimum time required to travel a distance $N$ with $M$ traffic lights, each having specific green and red durations. We are equipped with a "Nitrous Oxide" (Nitro) device that allows for instantaneous movement (teleportation) between traffic lights. However, this device cannot be used to cross zebra crossings at lights and requires a cooldown period equivalent to passing $K$ traffic lights manually before it can be used again. All traffic lights start their green cycle at $t=0$. If we arrive exactly when a light turns red, we must wait.

Problem Strategy

The problem can be modeled using dynamic programming. We need to track the minimum time to reach each traffic light while considering whether we arrived there via normal driving or by using the Nitro boost. A key observation from the problem constraints and sample cases is that "maximum speed $V$" acts as a multiplier where the time taken to drive a distance $D$ is $D \times V$. This is unconventional but necessary to match the provided sample logic.

To simplify the logic, we treat the final destination (the company) as an additional traffic light at distance $N$ that is always green (e.g., green duration 1, red duration 0).

Dynamic Programming State

We define a DP table dp[i][2] where:

• dp[i][0]: The minimum time to clear the $i$-th traffic light after arriving via normal driving from the previous light.
• dp[i][1]: The minimum time to clear the $i$-th traffic light after arriving via a Nitro boost from the previous light.

Transitions

For each light $i$ from 1 to $M+1$:

1. Driving Transition: To arrive at light $i$ by driving, we take the minimum time to have cleared light $i-1$ (either via driving or Nitro) and add the travel time $ (pos[i] - pos[i-1]) \times V $, then calculate any potential wait time at the red light.
2. Nitro Transition: To use the Nitro boost for the segment $(i-1) \to i$, we must satisfy the cooldown $K$. This means we look back to light $j = \max(0, i-K)$. We assume we were at light $j$ at its minimum time, and then we drove through all subsequent lights until $i-1$. Finally, we "teleport" the distance from $i-1$ to $i$.

Implementation

import java.util.Scanner;

public class NitroTrafficSolver {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        
        long totalDist = input.nextLong();
        int lightCount = input.nextInt();
        int cooldownK = input.nextInt();
        long speedMultiplier = input.nextLong();
        
        // We add one extra 'light' representing the destination
        int totalNodes = lightCount + 1;
        long[] locations = new long[totalNodes + 1];
        long[] greenTimes = new long[totalNodes + 1];
        long[] redTimes = new long[totalNodes + 1];
        
        for (int i = 1; i <= lightCount; i++) {
            locations[i] = input.nextLong() * speedMultiplier;
            greenTimes[i] = input.nextLong();
            redTimes[i] = input.nextLong();
        }
        
        // Company location as a permanent green light
        locations[totalNodes] = totalDist * speedMultiplier;
        greenTimes[totalNodes] = 1; 
        redTimes[totalNodes] = 0;

        // dp[i][0] -> Arrived at light i via driving
        // dp[i][1] -> Arrived at light i via Nitro
        long[][] dp = new long[totalNodes + 1][2];

        for (int i = 1; i <= totalNodes; i++) {
            // Case 1: Driving from the previous node
            long segmentDist = locations[i] - locations[i - 1];
            long driveTime1 = calculateTimeAtLight(dp[i - 1][0], segmentDist, greenTimes[i], redTimes[i]);
            long driveTime2 = calculateTimeAtLight(dp[i - 1][1], segmentDist, greenTimes[i], redTimes[i]);
            dp[i][0] = Math.min(driveTime1, driveTime2);

            // Case 2: Using Nitro for the last segment
            // Must have passed K lights manually since the last Nitro use
            int startNode = Math.max(0, i - cooldownK);
            long nitroStartTime = Math.min(dp[startNode][0], dp[startNode][1]);
            
            for (int j = startNode; j < i; j++) {
                long d = (j == i - 1) ? 0 : (locations[j + 1] - locations[j]);
                nitroStartTime = calculateTimeAtLight(nitroStartTime, d, greenTimes[j + 1], redTimes[j + 1]);
            }
            dp[i][1] = nitroStartTime;
        }

        System.out.println(Math.min(dp[totalNodes][0], dp[totalNodes][1]));
    }

    /**
     * Calculates the time when a vehicle clears a light.
     * @param start The time the vehicle cleared the previous light.
     * @param travel The time spent driving to the current light.
     * @param g Duration of green light.
     * @param r Duration of red light.
     * @return Time when the light turns green and the vehicle can pass.
     */
    private static long calculateTimeAtLight(long start, long travel, long g, long r) {
        long arrivalTime = start + travel;
        long cycle = g + r;
        long timeInCycle = arrivalTime % cycle;
        
        if (timeInCycle >= g) {
            // Arrived during red light, wait for next green
            return arrivalTime + (cycle - timeInCycle);
        }
        return arrivalTime;
    }
}

The time complexity of this approach is $O(M \times K)$, where $M$ is the number of traffic lights and $K$ is the cooldown period. Given that $M, K \le 1000$, the complexity $O(M^2)$ effectively handles the maximum constraints within the time limit. The space complexity is $O(M)$ to store the DP table and light configurations.