到达目的地的最少停留

Question

Class GasStation {
    int distanceToDestination;
    int availableGas;
}

给定三个参数g表示车辆初始油量，d表示到目的地的距离。和 gasStations 列表，其中每个 gasStation 的变量是 distanceToDestination，第二个是该站的 availableGas。如何计算到达目的地的最小停靠点？

g = 10 gallon,
d = 20 miles,
list of GasStation:
gasStations = [[15, 1], [14,10], [12,12]].

编辑：没有容量限制。

Answer 1

既然你没有提到它，我假设你需要 k 加仑汽油才能行驶 1 英里。如果总容量不是太大，这可以通过 DP 来解决。我已经概述了一个使用递归和记忆的解决方案。

gasStations  = [list of GasStations]
sort gasStations  by decreasing value of distanceToDestination if its not already sorted
k : gas required to travel 1 mile
maxNumberOfGasStation : maximum gas stations possible
maxPossibleCapacity : maximum gas that might be required for a trip
memo = [maxNumberOfGasStation][maxPossibleCapacity] filled up with -1

int f(idx, currentGas) {
    if (G[idx].distanceToDestination * k <= current_gas) {
        // You can reach destination using the gas you have left without filling any more
        return 0
    }
    if(idx == gasStations.length - 1) {
        // last station
        if (G[idx].distanceToDestination * k > current_gas + G[idx].availableGas) {
            // You cannot reach destination even if you fill up here
            return INT_MAX
        } else{
            return 1;
        }
    }   
    if(memo[idx][currentGas] != -1) return memo[idx][currentGas];

    // option 1: stop at this station
    int distBetweenStation = G[idx].distanceToDestination - G[idx+1].distanceToDestination
    int r1 = 1 + f(idx+1, min(currentGas + G[idx].availableGas, maxPossibleCapacity) - distBetweenStation * k)

    // option 2: don't stop at this station
    int r2 = f(idx+1, currentGas - distBetweenStation * k)

    // take minimum
    int r = min(r1, r2)

    memo[idx][currentGas] = r
    return r;
}

要接听电话f(0, g - (d - gasStations[0].distanceToDestination) * k)。时间复杂度为 O(maxNumberOfGasStation * maxPossibleCapacity)。如果有 capicity 限制，您可以简单地用它替换 maxPossibleCapacity。

到达目的地的最少停留

minimum stop to get to destination

algorithm

dynamic-programming

greedy