algorithm - 到达目的地的最短停留时间

Class GasStation {
    int distanceToDestination;
    int availableGas;
}

给定三个参数d g表示车辆初始气体量，d表示到目的地的距离。以及一个气体状态列表，其中对于每个气体状态变量是distanceToDestination，第二个变量是该站点的availableGas如何计算到达目的地的最小站点？

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

编辑：没有容量限制。

最佳答案

既然你没有提到，我想你需要k加仑的汽油才能行驶1英里。如果总容量不太大，这可以通过dp来解决。我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;
}

要获得答案，请致电f(0, g - (d - gasStations[0].distanceToDestination) * k)时间复杂度O(maxNumberOfGasStation * maxPossibleCapacity)。如果有一个capicity限制，您可以简单地用它替换maxPossibleCapacity。