我发现此代码使用蛮力机制解决背包问题(这主要是为了学习，因此无需指出动态更有效)。我得到了可以工作的代码，并且了解了大部分代码。最多。这是问题:

我注意到这两个条件，我不知道它们如何工作以及为什么在代码中-我知道它们至关重要，因为我进行的任何更改都会导致算法产生错误的结果:

// if bit not included then skip
if (((i >> j) & 1) != 1) continue;

// if bit match then add
if (((bestPosition >> j) & 1) == 1)
{
    include.Add(Items[j]);
}

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

namespace KnapSack2
{
    class BruteForce
    {
        public double Capacity { get; set; }
        public Item[] Items { get; set; }

        public Data Run()
        {
            int bestValue = 0;
            int bestPosition = 0;
            int size = Items.Length;

            var permutations = (long)Math.Pow(2,size);

            for (int i = 0; i<permutations; i++)
            {
                int total = 0;
                int weight = 0;
                for (int j = 0; j<size; j++)
                {
                    //jeżeli bit "not included" to omin"
                    if(((i>>j)&1)!=1)
                        continue;
                    total += Items[j].v;
                    weight += Items[j].w;
                }
                if (weight <= Capacity && total>bestValue)
                {
                    bestPosition = i;
                    bestValue = total;
                }
            }
            var include = new List<Item>();
            for (int j = 0; j<size; j++)
            {
                // jeżeli bit pasuje, wtedy dodaj
                if (((bestPosition>>j) & 1)==1)
                    include.Add(Items[j]);
            }
            return new Data { BestValue = bestValue, Include = include };

        }//End of Run


    }
}

var ks = new BruteForce
{
    Capacity = MaxWeight,
    Items = items.ToArray()
};

Data result = ks.Run();

最佳答案

此&是bitwise-AND

The bitwise-AND operator compares each bit of its first operand to the corresponding bit of its second operand. If both bits are 1, the corresponding result bit is set to 1. Otherwise, the corresponding result bit is set to 0.

The right-shift operator (>>) shifts its first operand right by the number of bits specified by its second operand.

if (((i >> j) & 1) != 1) continue;

int number = 5;

0000 0000 0000 0000 0000 0000 0000 0101

int shiftNumbersBitByOne = a >> 1;

0000 0000 0000 0000 0000 0000 0000 0010

What does exactly this operator ?

a =     0001 0100 1010 0001 1000 1010 1101 0011
b =     0010 1100 1111 0111 0011 1010 1011 0111
a & b = 0000 0100 1010 0001 0000 1010 1001 0011

0000 0000 0000 0000 0000 0000 0000 0001

When the result of (i >> j) & 1 would be 1?

The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a mass and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.

// This is the total weight limit.
public double Capacity { get; set; }

// This is an array with all the given items.
public Item[] Items { get; set; }

var permutations = (long)Math.Pow(2,size);

for (int i = 0; i<permutations; i++)
{ 
    // ...
}

for (int j = 0; j < size; j++)
{
    // Here you check if the item in position j
    // is included in the current combination.
    // If it is not, you go to the next value of the loop's variable
    // and you make the same check.
    if(((i>>j)&1)!=1)
        continue;

    // The j-th item is included in the current combination. 
    // Hence we add it's
    // value to the total value accumulator and it's weight to 
    // the total weight accumulator.
    total += Items[j].v;
    weight += Items[j].w;
}

bestPosition = i;
bestValue = total;

// The include is a list that will hold the items of the best combination.
var include = new List<Item>();

// We loop through all the available items
for (int j = 0; j<size; j++)
{
    // If the items is included in the best combination,
    // add this item to the include list.
    if (((bestPosition>>j) & 1)==1)
        include.Add(Items[j]);
}

关于c# - 背包-蛮力算法，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/29669259/