请注意，这是 C＃.NET 2.0 项目（ Linq不允许）所必需的。
我知道这里已经提出了非常类似的问题，我已经制作了一些工作代码（见下文）但仍然想知道如何在k和s条件下更快地使算法更快。
这是我迄今为止所学到的： 动态编程是查找ONE（不是所有）子集的最有效方法。如果我错了，请纠正我。有没有办法反复调用DP代码来生成更新的子集，直到包（重复设置）用完为止？
如果没有，那么有没有一种方法可以加快我下面的回溯递归算法，它确实产生了我需要的但是在O（2 ^ n）中运行，我认为，考虑到s和k？ / p>
这是我固定的数字包，不会改变，n = 114，数字范围从3到286：
int[] numbers = new int[]
{
7, 286, 200, 176, 120, 165, 206, 75, 129, 109,
123, 111, 43, 52, 99, 128, 111, 110, 98, 135,
112, 78, 118, 64, 77, 227, 93, 88, 69, 60,
34, 30, 73, 54, 45, 83, 182, 88, 75, 85,
54, 53, 89, 59, 37, 35, 38, 29, 18, 45,
60, 49, 62, 55, 78, 96, 29, 22, 24, 13,
14, 11, 11, 18, 12, 12, 30, 52, 52, 44,
28, 28, 20, 56, 40, 31, 50, 40, 46, 42,
29, 19, 36, 25, 22, 17, 19, 26, 30, 20,
15, 21, 11, 8, 8, 19, 5, 8, 8, 11,
11, 8, 3, 9, 5, 4, 7, 3, 6, 3,
5, 4, 5, 6
};
要求
空间限制为最大2-3GB，但时间应为O（n ^某事） （某物^ N）。
不得对行李进行分类，不得删除行李。
结果应该是匹配中数字的索引 子集，而不是数字本身（因为我们有重复）。
动态编程尝试
这是C＃动态编程版本，改编自stackoverflow.com上类似问题的答案：
using System;
using System.Collections.Generic;
namespace Utilities
{
public static class Combinations
{
private static Dictionary<int, bool> m_memo = new Dictionary<int, bool>();
private static Dictionary<int, KeyValuePair<int, int>> m_previous = new Dictionary<int, KeyValuePair<int, int>>();
static Combinations()
{
m_memo.Clear();
m_previous.Clear();
m_memo[0] = true;
m_previous[0] = new KeyValuePair<int, int>(-1, 0);
}
public static bool FindSubset(IList<int> set, int sum)
{
//m_memo.Clear();
//m_previous.Clear();
//m_memo[0] = true;
//m_previous[0] = new KeyValuePair<int, int>(-1, 0);
for (int i = 0; i < set.Count; ++i)
{
int num = set[i];
for (int s = sum; s >= num; --s)
{
if (m_memo.ContainsKey(s - num) && m_memo[s - num] == true)
{
m_memo[s] = true;
if (!m_previous.ContainsKey(s))
{
m_previous[s] = new KeyValuePair<int, int>(i, num);
}
}
}
}
return m_memo.ContainsKey(sum) && m_memo[sum];
}
public static IEnumerable<int> GetLastIndex(int sum)
{
while (m_previous[sum].Key != -1)
{
yield return m_previous[sum].Key;
sum -= m_previous[sum].Value;
}
}
public static void SubsetSumMain(string[] args)
{
int[] numbers = new int[]
{
7, 286, 200, 176, 120, 165, 206, 75, 129, 109,
123, 111, 43, 52, 99, 128, 111, 110, 98, 135,
112, 78, 118, 64, 77, 227, 93, 88, 69, 60,
34, 30, 73, 54, 45, 83, 182, 88, 75, 85,
54, 53, 89, 59, 37, 35, 38, 29, 18, 45,
60, 49, 62, 55, 78, 96, 29, 22, 24, 13,
14, 11, 11, 18, 12, 12, 30, 52, 52, 44,
28, 28, 20, 56, 40, 31, 50, 40, 46, 42,
29, 19, 36, 25, 22, 17, 19, 26, 30, 20,
15, 21, 11, 8, 8, 19, 5, 8, 8, 11,
11, 8, 3, 9, 5, 4, 7, 3, 6, 3,
5, 4, 5, 6
};
int sum = 400;
//int size = 4; // don't know to use in dynamic programming
// call dynamic programming
if (Numbers.FindSubset(numbers, sum))
{
foreach (int index in Numbers.GetLastIndex(sum))
{
Console.Write((index + 1) + "." + numbers[index] + "\t");
}
Console.WriteLine();
}
Console.WriteLine();
Console.ReadKey();
}
}
}
递归编程尝试
这是C＃递归编程版本，改编自stackoverflow.com上类似问题的答案：
using System;
using System.Collections.Generic;
namespace Utilities
{
public static class Combinations
{
private static int s_count = 0;
public static int CountSubsets(int[] numbers, int index, int current, int sum, int size, List<int> result)
{
if ((numbers.Length <= index) || (current > sum)) return 0;
if (result == null) result = new List<int>();
List<int> temp = new List<int>(result);
if (current + numbers[index] == sum)
{
temp.Add(index);
if ((size == 0) || (temp.Count == size))
{
s_count++;
}
}
else if (current + numbers[index] < sum)
{
temp.Add(index);
CountSubsets(numbers, index + 1, current + numbers[index], sum, size, temp);
}
CountSubsets(numbers, index + 1, current, sum, size, result);
return s_count;
}
private static List<List<int>> m_subsets = new List<List<int>>();
public static List<List<int>> FindSubsets(int[] numbers, int index, int current, int sum, int size, List<int> result)
{
if ((numbers.Length <= index) || (current > sum)) return m_subsets;
if (result == null) result = new List<int>();
List<int> temp = new List<int>(result);
if (current + numbers[index] == sum)
{
temp.Add(index);
if ((size == 0) || (temp.Count == size))
{
m_subsets.Add(temp);
}
}
else if (current + numbers[index] < sum)
{
temp.Add(index);
FindSubsets(numbers, index + 1, current + numbers[index], sum, size, temp);
}
FindSubsets(numbers, index + 1, current, sum, size, result);
return m_subsets;
}
public static void SubsetSumMain(string[] args)
{
int[] numbers = new int[]
{
7, 286, 200, 176, 120, 165, 206, 75, 129, 109,
123, 111, 43, 52, 99, 128, 111, 110, 98, 135,
112, 78, 118, 64, 77, 227, 93, 88, 69, 60,
34, 30, 73, 54, 45, 83, 182, 88, 75, 85,
54, 53, 89, 59, 37, 35, 38, 29, 18, 45,
60, 49, 62, 55, 78, 96, 29, 22, 24, 13,
14, 11, 11, 18, 12, 12, 30, 52, 52, 44,
28, 28, 20, 56, 40, 31, 50, 40, 46, 42,
29, 19, 36, 25, 22, 17, 19, 26, 30, 20,
15, 21, 11, 8, 8, 19, 5, 8, 8, 11,
11, 8, 3, 9, 5, 4, 7, 3, 6, 3,
5, 4, 5, 6
};
int sum = 17;
int size = 2;
// call backtracking recursive programming
Console.WriteLine("CountSubsets");
int count = Numbers.CountSubsets(numbers, 0, 0, sum, size, null);
Console.WriteLine("Count = " + count);
Console.WriteLine();
// call backtracking recursive programming
Console.WriteLine("FindSubsets");
List<List<int>> subsets = Numbers.FindSubsets(numbers, 0, 0, sum, size, null);
for (int i = 0; i < subsets.Count; i++)
{
if (subsets[i] != null)
{
Console.Write((i + 1).ToString() + ":\t");
for (int j = 0; j < subsets[i].Count; j++)
{
int index = subsets[i][j];
Console.Write((index + 1) + "." + numbers[index] + " ");
}
Console.WriteLine();
}
}
Console.WriteLine("Count = " + subsets.Count);
Console.ReadKey();
}
}
}
请让我知道如何将动态编程版本限制为大小为k的子集，如果我可以重复调用它，那么它会在每次调用时返回不同的子集，直到没有更多匹配的子集。
此外，我不确定在何处初始化DP算法的备忘录。我是在静态构造函数中完成的，它在访问任何方法时自动运行。这是正确的初始化位置还是需要移动到FindSunset（）方法内部[注释掉]？
至于递归版本，它是回溯吗？我们怎样才能加快速度。它工作正常，考虑到k和s，但效率很低。
让这个主题成为所有C＃SubsetSum相关问题的母亲！
答案 0 :(得分：0)
只需搜索大小为K的所有组合，并检查每个组合是否满足条件。
适合您情况的k组合的最快算法是：
for (var i1 = 0; i1 <= n; i1++)
{
for (var i2 = i1 + 1; i2 <= n; i2++)
{
for (var i3 = i2 + 1; i3 <= n; i3++)
{
...
for (var ik = iOneBeforeK + 1; ik <= n; ik++)
{
if (arr[i1] + arr[i2] + ... + arr[ik] == sum)
{
// this is a valid subset
}
}
}
}
}
但是你在谈论数字并总结它们，这意味着你可以用更聪明的算法来制作截止数据。
由于所有数字都是正数，因此您知道如果单个数字足够大，则无法向其添加任何更多正数并将其总和为s。鉴于s=6和k=4，搜索中包含的最高数字为s-k+1=3。 3+1+1+1是k个数字，1是您可能的最低数字，总计为6。任何高于3的数字都不能添加3个其他正数，并且总和<= 6。
但请等一下，您的最低可能值不是1，而是3。那甚至更好。因此，假设k=10，n=60，min=3。 ＆＃34;最高数字情景＆＃34;是x+min(k-1)=n - ＆gt; x=60-3*9=33。因此，即使考虑的最高数字也是33。
这会减少要考虑的数组中的相关数量，因此会减少要查找的组合数量。
但它变得更好。假设k=10，n=60，min=3。数组中的第一个数字恰好是20，因此它是相关的并且应该被检查。让我们找到包含那个20的相关子集：
一个新的＆＃34;难题＆＃34;出现！ k=10-1，n=60-20，min=3。你现在可以从子插座中切断许多数字，并且每一步都会一次又一次地切断它。
通过计算子puzzle中k-1最小数字的平均值并将其用作min，可以进一步改善这一点。
通过预先计算子项目k中的[0..n]最低平均数，以及k-1中[1..n]最低数字k-2和{{}，可以进一步提高这一点。 1}} subpuzzle [2..n]中的平均值最低等等，并使用它们而不是在每个子项目评估中反复重新计算相同的内容。
答案 1 :(得分：0)
可以通过与背包问题类似的解决方案来解决
dp [i] [j] [k] =使用前“ i”个元素的总和等于j的k个子集的数量
dp [i] [j] [k] = dp [i-1] [j] [k] + dp [i-1] [j-a [i]] [k-1]
是dp的更新（使用ith元素还是不使用ith元素）
for(int i=0;i<=n;i++) dp[i][0][0]=1;
for(int i=1;i<=n;i++){
for(int j=0;j<=w;j++){
for(int k=1;k<=i;k++){
dp[i][j][k]=dp[i-1][j][k] ;
if(j>=a[i-1]){
dp[i][j][k]+=dp[i-1][j-a[i-1]][k-1];
}
}
}
}
答案 2 :(得分：0)
有多种解决方案，但是没有人展示如何使用动态编程来找到答案。
关键是使用动态编程来建立一个数据结构，以便以后可以从中找到所有解决方案。
除了请求的功能外，我还收集了有关有多少个解决方案的信息，并写了FindSolution(node, position)以返回从position开始的位置node的解决方案，而没有计算其余的结果。如果您想全部使用它们，那么使用该功能将效率很低。但是，例如，使用此功能，可以计算十亿分之一的方式来表示10000作为20个质数的总和。使用给定的其他方法，这是不可行的。
using System;
using System.Collections.Generic;
public class SubsetSum
{
public class SolutionNode<T>
{
// The index we found the value at
public int Index {get; set;}
// The value we add for this solution
public T Value {get; set;}
// How many solutions we have found.
public int Count {get; set;}
// The rest of this solution.
public SolutionNode<T> Tail {get; set;}
// The next solution.
public SolutionNode<T> Next {get; set;}
}
// This uses dynamic programming to create a summary of all solutions.
public static SolutionNode<int> FindSolution(int[] numbers, int target, int subsetSize)
{
// By how many are in our list, by what they sum to, what SolutionNode<int> has our answer?
List<Dictionary<int, SolutionNode<int>>> solutionOf = new List<Dictionary<int, SolutionNode<int>>>();
// Initialize empty solutions.
for (int i = 0; i <= subsetSize; i++)
{
solutionOf.Add(new Dictionary<int, SolutionNode<int>>());
}
// We discover from the last number in the list forward.
// So discovering from the last index forward makes them ordered.
for (int i = numbers.Length - 1; -1 < i; i--)
{
int number = numbers[i];
// Descending here so we don't touch solutionOf[j-1] until after we have solutionOf[j] updated.
for (int j = subsetSize; 0 < j; j--)
{
// All previously found sums with j entries
Dictionary<int, SolutionNode<int>> after = solutionOf[j];
// All previously found sums with j-1 entries
Dictionary<int, SolutionNode<int>> before = solutionOf[j-1];
foreach (KeyValuePair<int, SolutionNode<int>> pair in before)
{
SolutionNode<int> newSolution = new SolutionNode<int>();
int newSum = pair.Key + number;
newSolution.Index = i;
newSolution.Value = number;
newSolution.Count = pair.Value.Count;
newSolution.Tail = pair.Value;
if (after.ContainsKey(newSum))
{
newSolution.Next = after[newSum];
newSolution.Count = pair.Value.Count + after[newSum].Count;
}
after[newSum] = newSolution;
}
// And special case empty set.
if (1 == j)
{
SolutionNode<int> newSolution = new SolutionNode<int>();
newSolution.Index = i;
newSolution.Value = number;
newSolution.Count = 1;
if (after.ContainsKey(number))
{
newSolution.Next = after[number];
newSolution.Count = after[number].Count;
}
after[number] = newSolution;
}
}
}
// Return what we found.
try
{
return solutionOf[subsetSize][target];
}
catch
{
throw new Exception("No solutions found");
}
}
// The function we were asked for.
public static IEnumerable<List<int>> ListSolutions (SolutionNode<int> node)
{
List<int> solution = new List<int>();
List<SolutionNode<int>> solutionPath = new List<SolutionNode<int>>();
// Initialize with starting information.
solution.Add(0); // This will be removed when we get node
solutionPath.Add(node); // This will be our start.
while (0 < solutionPath.Count)
{
// Erase the tail of our previous solution
solution.RemoveAt(solution.Count - 1);
// Pick up our next.
SolutionNode<int> current = solutionPath[solutionPath.Count - 1];
solutionPath.RemoveAt(solutionPath.Count - 1);
while (current != null)
{
solution.Add(current.Index);
solutionPath.Add(current.Next);
if (current.Tail == null)
{
yield return solution;
}
current = current.Tail;
}
}
}
// And for fun, a function that dynamic programming makes easy - return any one of them!
public static List<int> FindSolution(SolutionNode<int> node, int position)
{
// Switch to counting from the end.
position = node.Count - position - 1;
List<int> solution = new List<int>();
while (node != null)
{
while (node.Next != null && position < node.Next.Count)
{
node = node.Next;
}
solution.Add(node.Index);
node = node.Tail;
}
return solution;
}
public static void Main(string[] args)
{
SolutionNode<int> solution = FindSolution(
new[]{
7, 286, 200, 176, 120, 165, 206, 75, 129, 109,
123, 111, 43, 52, 99, 128, 111, 110, 98, 135,
112, 78, 118, 64, 77, 227, 93, 88, 69, 60,
34, 30, 73, 54, 45, 83, 182, 88, 75, 85,
54, 53, 89, 59, 37, 35, 38, 29, 18, 45,
60, 49, 62, 55, 78, 96, 29, 22, 24, 13,
14, 11, 11, 18, 12, 12, 30, 52, 52, 44,
28, 28, 20, 56, 40, 31, 50, 40, 46, 42,
29, 19, 36, 25, 22, 17, 19, 26, 30, 20,
15, 21, 11, 8, 8, 19, 5, 8, 8, 11,
11, 8, 3, 9, 5, 4, 7, 3, 6, 3,
5, 4, 5, 6}
, 400, 4);
IEnumerable<List<int>> listing = ListSolutions(solution);
foreach (List<int> sum in listing)
{
Console.WriteLine ("solution {0}", string.Join(", ", sum.ToArray()));
}
}
}
顺便说一句，这是我第一次尝试编写C＃。太痛苦了。