N个城市之间可能的最小旅行成本

在一条笔直的道路上有N个城市，每个城市相隔1 个单位的距离。您必须乘坐公共汽车到达第(N + 1)^个城市。第i^个城市旅行1 个单位的距离将花费C[i]美元。换句话说，从第i^个城市到第j^个城市的旅行成本是abs(i – j ) * C[i]美元。任务是找到从城市1到城市(N + 1)即超出最后一个城市的最低成本。
例子：

Input: C[] = {3, 5, 4} 
Output: 9 
The bus boarded from the first city has the minimum 
cost of all so it will be used to travel (N + 1) unit.
Input: C[] = {4, 7, 8, 3, 4} 
Output: 18 
Board the bus at the first city then change 
the bus at the fourth city. 
(3 * 4) + (2 * 3) = 12 + 6 = 18 
 

方法：方法很简单，乘坐目前成本最低的巴士即可。每当发现成本更低的巴士时，请更换该城市的巴士。以下是解决步骤：

1. 从第一个城市开始，成本为C[1] 。
2. 前往下一个城市，直到找到比前一个城市（我们正在旅行的城市，假设城市i ）花费更少的城市j 。
3. 计算成本为abs(j – i) * C[i]并将其添加到目前的总成本中。
4. 重复前面的步骤，直到遍历所有城市。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimum cost to
// travel from the first city to the last
int minCost(vector& cost, int n)
{
 
    // To store the total cost
    int totalCost = 0;
 
    // Start from the first city
    int boardingBus = 0;
 
    for (int i = 1; i < n; i++) {
 
        // If found any city with cost less than
        // that of the previous boarded
        // bus then change the bus
        if (cost[boardingBus] > cost[i]) {
 
            // Calculate the cost to travel from
            // the currently boarded bus
            // till the current city
            totalCost += ((i - boardingBus) * cost[boardingBus]);
 
            // Update the currently boarded bus
            boardingBus = i;
        }
    }
 
    // Finally calculate the cost for the
    // last boarding bus till the (N + 1)th city
    totalCost += ((n - boardingBus) * cost[boardingBus]);
    return totalCost;
}
 
// Driver code
int main()
{
    vector cost{ 4, 7, 8, 3, 4 };
    int n = cost.size();
 
    cout << minCost(cost, n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the minimum cost to
// travel from the first city to the last
static int minCost(int []cost, int n)
{
 
    // To store the total cost
    int totalCost = 0;
 
    // Start from the first city
    int boardingBus = 0;
 
    for (int i = 1; i < n; i++)
    {
 
        // If found any city with cost less than
        // that of the previous boarded
        // bus then change the bus
        if (cost[boardingBus] > cost[i])
        {
 
            // Calculate the cost to travel from
            // the currently boarded bus
            // till the current city
            totalCost += ((i - boardingBus) * cost[boardingBus]);
 
            // Update the currently boarded bus
            boardingBus = i;
        }
    }
 
    // Finally calculate the cost for the
    // last boarding bus till the (N + 1)th city
    totalCost += ((n - boardingBus) * cost[boardingBus]);
    return totalCost;
}
 
// Driver code
public static void main(String[] args)
{
    int []cost = { 4, 7, 8, 3, 4 };
    int n = cost.length;
 
    System.out.print(minCost(cost, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
 
# Function to return the minimum cost to
# travel from the first city to the last
def minCost(cost, n):
 
    # To store the total cost
    totalCost = 0
 
    # Start from the first city
    boardingBus = 0
 
    for i in range(1, n):
 
        # If found any city with cost less than
        # that of the previous boarded
        # bus then change the bus
        if (cost[boardingBus] > cost[i]):
 
            # Calculate the cost to travel from
            # the currently boarded bus
            # till the current city
            totalCost += ((i - boardingBus) *
                          cost[boardingBus])
 
            # Update the currently boarded bus
            boardingBus = i
 
    # Finally calculate the cost for the
    # last boarding bus till the (N + 1)th city
    totalCost += ((n - boardingBus) *
                  cost[boardingBus])
    return totalCost
 
# Driver code
cost = [ 4, 7, 8, 3, 4]
n = len(cost)
 
print(minCost(cost, n))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the minimum cost to
// travel from the first city to the last
static int minCost(int []cost, int n)
{
 
    // To store the total cost
    int totalCost = 0;
 
    // Start from the first city
    int boardingBus = 0;
 
    for (int i = 1; i < n; i++)
    {
 
        // If found any city with cost less than
        // that of the previous boarded
        // bus then change the bus
        if (cost[boardingBus] > cost[i])
        {
 
            // Calculate the cost to travel from
            // the currently boarded bus
            // till the current city
            totalCost += ((i - boardingBus) *
                          cost[boardingBus]);
 
            // Update the currently boarded bus
            boardingBus = i;
        }
    }
 
    // Finally calculate the cost for the
    // last boarding bus till the (N + 1)th city
    totalCost += ((n - boardingBus) *
                  cost[boardingBus]);
    return totalCost;
}
 
// Driver code
public static void Main(String[] args)
{
    int []cost = { 4, 7, 8, 3, 4 };
    int n = cost.Length;
 
    Console.Write(minCost(cost, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

// javascript implementation of the approach   
// Function to return the minimum cost to
    // travel from the first city to the last
    function minCost(cost , n) {
 
        // To store the total cost
        var totalCost = 0;
 
        // Start from the first city
        var boardingBus = 0;
 
        for (i = 1; i < n; i++) {
 
            // If found any city with cost less than
            // that of the previous boarded
            // bus then change the bus
            if (cost[boardingBus] > cost[i]) {
 
                // Calculate the cost to travel from
                // the currently boarded bus
                // till the current city
                totalCost += ((i - boardingBus) * cost[boardingBus]);
 
                // Update the currently boarded bus
                boardingBus = i;
            }
        }
 
        // Finally calculate the cost for the
        // last boarding bus till the (N + 1)th city
        totalCost += ((n - boardingBus) * cost[boardingBus]);
        return totalCost;
    }
 
    // Driver code
     
        var cost = [ 4, 7, 8, 3, 4 ];
        var n = cost.length;
 
        document.write(minCost(cost, n));
 
// This code contributed by umadevi9616

输出：

18

时间复杂度： O(N)
辅助空间： O(1)

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。