谦卑挑战

Question

为了提高我的编程技能，或者说我缺乏编程技巧，我正试图完成这一谦逊的挑战。挑战的细节在这里。

房间里有N根绳子和N个重量。每根绳子都与一个重量相连(只在一端)，而且每根绳子都有一个特殊的耐用性−，这是它可以挂起的最大重量。还有一个挂钩，挂在天花板上。这些绳子可以在没有重量的情况下绑在钩子上。这些绳索也可以连接到其他重量上，也就是说，绳子和重物可以在一条链子中相互连接。如果与绳子直接或间接连接的重量之和大于其耐久性，绳子就会断裂。 我们知道我们想要附加N根绳子的顺序。更准确地说，我们知道绳子的参数(耐用性和重量)和每个附件的位置。在三个长度为N的零索引数组A、B、C中给出了耐久度、权重和位置。对于每一个i (0≤iA= 4,3,1 B= 2,2,1 C= -1,0,1 函数应该返回2，就好像我们附加了第三根绳子，那么一根绳子就会断掉，因为重量之和大于它的耐久性(2 +2+1=5和5> 4)。

下面是我尝试的解决方案。我有一个名为add_weights的助手函数，它返回True，如果添加最新的绳子不会导致任何其他绳子断裂，否则会出错。

def add_weights(A, ancestors, weights, weight, rope):
    #add weight(int) to rope and ancestors
    if (rope == -1):
        return (True)
    else:
        weights[rope] += weight
        if (A[rope] < weights[rope]):
            print "Rope that breaks", rope
            return False
        ancestor = ancestors[rope]
        print ancestor
        add_weights(A, ancestors, weights, weight, ancestor)




def solution(A, B, C):
    # write your code in Python 2.7
    weights = {}
    ancestors = {}
    for i in range(len(B)):
        weights[i] = B[i]
    for i in range(len(C)):
        #attaching rope i to rope x
        x = C[i]
        ancestors[i] = x
        broke = add_weights(A, ancestors, weights, B[i], x)
        if (not broke):
            return i
    return len(C)

问题是在函数解决方案中的for循环的第二次迭代中(当我尝试添加rope 1时)，当我可以清楚地看到add_weights返回True时，变量中断会以某种方式计算为零。我也用调试器对它进行了测试，所以我不完全确定发生了什么。欢迎任何帮助。

Answer 1

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace Exercise5_DurablityofRopes
{
    class Program
    {
           static int[] A_Durability = new int[] { 15, 6, 2,3,1 };
           static int[] B_Weight = new int[] { 2, 1, 2,3,1 };
           static int[] C_Position = new int[] { -1, 0, 1 ,2,3};
           static int no_of_ropes_attached = 0;
           static int maxropeattached = 0;
           static void Main(string[] args)
        {
           // first position cannot necessarily be -1 hence Checking for each position How many maximum ropes can be attached
            for (int i = 0; i <= C_Position.Length - 1; i++)
            {
                int[] Copy_Durability = new int[A_Durability.Length];
                for (int l = 0; l <= C_Position.Length - 1; l++)
                {
                    Copy_Durability[l] = A_Durability[l];
                }
                AddNextRope(i, B_Weight[i], Copy_Durability);          
                Console.WriteLine("Total Number of ropes can be attached to " + C_Position[i] + " ropes are" + no_of_ropes_attached);
                if (no_of_ropes_attached>=maxropeattached)
                {
                    maxropeattached = no_of_ropes_attached;
                }
                no_of_ropes_attached = 0;
            }

            Console.WriteLine("Total Number of ropes can be attached is  " + maxropeattached);
            Console.ReadKey();
        }

        private static void AddNextRope(int currentRopePosition,int newWeight,int[] Durability)
        {
            if (currentRopePosition == C_Position.Length - 1) return;
            // decrease same amount of weight from all ansestors from their durability and check if any of them breaks (durability <0) with new weight added
            for (int k = currentRopePosition; k != 0; k--)
            {
                Durability[k] = Durability[k] - newWeight;
                if(Durability[k]<0)
                {
                    return;
                }
            }
            no_of_ropes_attached = no_of_ropes_attached + 1; 

            for (int i = 0; i <= C_Position.Length - 1; i++)
                {
                    if (C_Position[i] == C_Position[currentRopePosition] + 1)
                    {
                        if (A_Durability[i] > B_Weight[i])
                        {
                            AddNextRope(i, B_Weight[i], Durability);
                        }
                        else
                        {
                            return;
                        }
                    }
                }
        }
    }
}

Answer 2

这是我的第一个解决方案O(N**2)，它简单地运行锻造绳索和减轻每个连接的重量。

  def solution(A, B, C):
    num_ropes = len(C)

    for i in range(num_ropes):
      node = i
      while(node != -1):
        A[node] -= B[i]
        if A[node] < 0:
          return i
        node = C[node]
    return num_ropes

这是O(N*log(N))的解决方案，其原理是，它不是去处理所有的绳子，而是使用具有一种张力的变形组。

  def solution(A, B, C):
    num_ropes = len(C)

    for i in range(num_ropes):
      A[i] -= B[i]
      if A[i] < 0:
        return i

      pre_node = C[i]
      while pre_node != -1:
        A[pre_node] -= B[i]
        if A[pre_node] < 0:
          return i

        if A[pre_node] <= A[i]:
          C[i] = pre_node
          A[i] = A[pre_node]
        pre_node = C[pre_node]

    return num_ropes

它可以使用最后一部分或在https://app.codility.com/programmers/challenges/sulphur2014/上进行测试。

  if __name__ == '__main__':
    A = [5, 3, 6, 3, 3]
    B = [2, 3, 1, 1, 2]
    C = [-1, 0, -1, 0, 3]
    assert solution(A, B, C) == 3

    A = [4, 3, 1]
    B = [2, 2, 1]
    C = [-1, 0, 1]
    assert solution(A, B, C) == 2

Answer 3

我没有时间将其转换为Python，但是通过查看这些JS代码，您可以有一个想法：

function keysOrderedByValues(array) {
    var result = [];
    for (var i = 0; i < array.length; i++) {
        result[array[i] + 1] = i;
    }
    return result;
}

function findFirstPositiveElem(array) {
    for (var i = 0; i < array.length; i++) {
        if (array[i]>0) {
            return i;
        }
    }
    return null;
}

function accumulativeArray(array) {
    var result = [];
    var sum = 0;
    for (var i = 0; i < array.length; i++) {
        sum = sum + array[i];
        result.push(sum);
    }
    return result;
}

Array.prototype.indexedByArray = function(array) {
    result = [];
    for (var i = 0; i < array.length; i++) {
        result.push(this[array[i]])
    }
    return result;
}

function test(ds, ws, ords) {
    return findFirstPositiveElem(ws.indexedByArray(keysOrderedByValues(ords)), ds);
}

console.log(test([4,3,1], [2,2,1], [-1,0,1]))