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并且我已经生成了一些工作代码(见下文)但是仍然想知道如何在k和s条件下更快地使算法更快.
这是我到目前为止所学到的：
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？
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 ^某事物)
(东西^ N).
>不得对行李进行分类,不得删除重复行李.
>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,我不知道在何处初始化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[注释掉]内？
至于递归版本,它是回溯吗？我们怎样才能加快速度.它正常工作并考虑k和s但完全没有效率.
让我们将这个主题作为所有C#SubsetSum相关问题的母亲！
最佳答案 我之前的回答是关于切断要检查的组合数量的原则.但是,一旦对数组进行排序,这可以得到显着改善.原理类似,但由于解决方案完全不同,我决定将其单独回答.
我小心翼翼地只使用.Net Framework 2.0功能.可能稍后添加一个视觉解释,但评论应该足够了.
class Puzzle
{
private readonly int[] _tailSums;
public readonly SubsetElement[] Elements;
public readonly int N;
public Puzzle(int[] numbers)
{
// Set N and make Elements array
// (to remember the original index of each element)
this.N = numbers.Length;
this.Elements = new SubsetElement[this.N];
for (var i = 0; i < this.N; i++)
{
this.Elements[i] = new SubsetElement(numbers[i], i);
}
// Sort Elements descendingly by their Number value
Array.Sort(this.Elements, (a, b) => b.Number.CompareTo(a.Number));
// Save tail-sums to allow immediate access by index
// Allow immedate calculation by index = N, to sum 0
this._tailSums = new int[this.N + 1];
var sum = 0;
for (var i = this.N - 1; i >= 0; i--)
{
this._tailSums[i] = sum += this.Elements[i].Number;
}
}
public void Solve(int s, Action<SubsetElement[]> callback)
{
for (var k = 1; k <= this.N; k++)
this.Solve(k, s, callback);
}
public void Solve(int k, int s, Action<SubsetElement[]> callback)
{
this.ScanSubsets(0, k, s, new List<SubsetElement>(), callback);
}
private void ScanSubsets(int startIndex, int k, int s,
List<SubsetElement> subset, Action<SubsetElement[]> cb)
{
// No more numbers to add, and current subset is guranteed to be valid
if (k == 0)
{
// Callback with current subset
cb(subset.ToArray());
return;
}
// Sum the smallest k elements
var minSubsetStartIndex = this.N - k;
var minSum = this._tailSums[minSubssetStartIndex];
// Smallest possible sum is greater than wanted sum,
// so a valid subset cannot be found
if (minSum > s)
{
return;
}
// Find largest number that satisfies the condition
// that a valid subset can be found
minSum -= this.Elements[minSubsetStartIndex].Number;
// But remember the last index that satisfies the condition
var minSubsetEndIndex = minSubsetStartIndex;
while (minSubsetStartIndex > startIndex &&
minSum + this.Elements[minSubsetStartIndex - 1].Number <= s)
{
minSubsetStartIndex--;
}
// Find the first number in the sorted sequence that is
// the largest number we just found (in case of duplicates)
while (minSubsetStartIndex > startIndex &&
Elements[minSubsetStartIndex] == Elements[minSubsetStartIndex - 1])
{
minSubsetStartIndex--;
}
// [minSubsetStartIndex .. maxSubsetEndIndex] is the
// full range we must check in recursion
for (var subsetStartIndex = minSubsetStartIndex;
subsetStartIndex <= minSubsetEndIndex;
subsetStartIndex++)
{
// Find the largest possible sum, which is the sum of the
// k first elements, starting at current subsetStartIndex
var maxSum = this._tailSums[subsetStartIndex] -
this._tailSums[subsetStartIndex + k];
// The largest possible sum is less than the wanted sum,
// so a valid subset cannot be found
if (maxSum < s)
{
return;
}
// Add current number to the subset
var x = this.Elements[subsetStartIndex];
subset.Add(x);
// Recurse through the sub-problem to the right
this.ScanSubsets(subsetStartIndex + 1, k - 1, s - x.Number, subset, cb);
// Remove current number and continue loop
subset.RemoveAt(subset.Count - 1);
}
}
public sealed class SubsetElement
{
public readonly int Number;
public readonly int Index;
public SubsetElement(int number, int index)
{
this.Number = number;
this.Index = index;
}
public override string ToString()
{
return string.Format("{0}({1})", this.Number, this.Index);
}
}
}
使用和性能测试：
private static void Main()
{
var sw = Stopwatch.StartNew();
var puzzle = new Puzzle(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
});
puzzle.Solve(2, 17, PuzzleOnSubsetFound);
sw.Stop();
Console.WriteLine("Subsets found: " + _subsetsCount);
Console.WriteLine(sw.Elapsed);
}
private static int _subsetsCount;
private static void PuzzleOnSubsetFound(Puzzle.SubsetElement[] subset)
{
_subsetsCount++;
return; // Skip prints when speed-testing
foreach (var el in subset)
{
Console.Write(el.ToString());
Console.Write(" ");
}
Console.WriteLine();
}
输出：
每一行都是找到的子集,()中的数字是子集中使用的数字的原始索引
14(60) 3(107)
14(60) 3(109)
14(60) 3(102)
13(59) 4(105)
13(59) 4(111)
12(64) 5(96)
12(64) 5(104)
12(64) 5(112)
12(64) 5(110)
12(65) 5(96)
12(65) 5(104)
12(65) 5(112)
12(65) 5(110)
11(100) 6(108)
11(100) 6(113)
11(61) 6(108)
11(61) 6(113)
11(92) 6(108)
11(92) 6(113)
11(62) 6(108)
11(62) 6(113)
11(99) 6(108)
11(99) 6(113)
9(103) 8(93)
9(103) 8(94)
9(103) 8(97)
9(103) 8(98)
9(103) 8(101)
Subsets found: 28
00:00:00.0017020 (measured when no printing is performed)
k越高,可以产生越多的截止.这是当你看到主要的性能差异时.当s变高时,您当前的代码(递归版本)的执行速度也会明显变慢.
With
k=5,s=60
Your current code: found 45070 subsets in 2.8602923 seconds
My code: found 45070 subsets in 0.0116727 seconds
That is 99.6% speed improvement
