到达目的地的最小停留

Question

.

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

Class GasStation {
    int distanceToDestination;
    int availableGas;
}

编辑：没有容量限制。

Answer 1

由于你没有提到它，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，可以通过DP解决这个问题。我已经概述了使用递归和memoization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maxPossibleCapacity替换它。