集合覆盖问题

Table of content

集合覆盖算法
实现

集合覆盖算法为许多现实世界中的资源分配问题提供了解决方案。例如，考虑一家航空公司为每架飞机分配机组人员，以确保他们有足够的人员来满足旅程的要求。他们会考虑航班时间、持续时间、中途停留以及机组人员的可用性，以便将他们分配到航班。这就是集合覆盖算法发挥作用的地方。

给定一个包含一些元素的全集 U，所有这些元素都被分成子集。将这些子集的集合视为 S = {S₁, S₂, S₃, S₄... S_n}，集合覆盖算法找到最少的子集数量，使得它们覆盖全集中的所有元素。

如上图所示，点表示存在于全集 U 中的元素，这些元素被分成不同的集合 S = {S₁, S₂, S₃, S₄, S₅, S₆}。为了覆盖所有元素，需要选择的最小集合数将是最优输出 = {S₁, S₂, S₃}。

集合覆盖算法

集合覆盖算法将集合的集合作为输入，并返回包含所有全集元素所需的最小集合数。

集合覆盖问题是一个 NP-Hard 问题，并且是一个 2-逼近贪心算法。

算法

步骤 1 - 初始化 Output = {}，其中 Output 表示输出元素集。

步骤 2 - 当 Output 集不包含全集中的所有元素时，执行以下操作：

使用公式 $\frac{Cost\left ( S_{i} \right )}{S_{i}-Output}$ 查找全集中的每个子集的成本效益。
找到每次迭代中成本效益最低的子集。将子集添加到 Output 集。

步骤 3 - 重复步骤 2，直到宇宙中没有剩余元素。达到的输出是最终的 Output 集。

伪代码

APPROX-GREEDY-SET_COVER(X, S)
   U = X
   OUTPUT = ф
   while U ≠ ф
      select S_i Є S which has maximum |S_i∩U|
   U = U – S
   OUTPUT = OUTPUT∪ {S_i}
return OUTPUT

分析

假设元素总数等于集合总数（|X| = |S|），则代码运行时间为 O(|X|³)

示例

让我们来看一个更详细地描述集合覆盖问题近似算法的例子

S₁ = {1, 2, 3, 4}                cost(S₁) = 5
S₂ = {2, 4, 5, 8, 10}            cost(S₂) = 10
S₃ = {1, 3, 5, 7, 9, 11, 13}     cost(S₃) = 20
S₄ = {4, 8, 12, 16, 20}          cost(S₄) = 12
S₅ = {5, 6, 7, 8, 9}             cost(S₅) = 15

步骤 1

输出集 Output = ф

找到输出集中没有元素时每个集合的成本效益，

S₁ = cost(S₁) / (S₁ – Output) = 5 / (4 – 0)
S₂ = cost(S₂) / (S₂ – Output) = 10 / (5 – 0)
S₃ = cost(S₃) / (S₃ – Output) = 20 / (7 – 0)
S₄ = cost(S₄) / (S₄ – Output) = 12 / (5 – 0)
S₅ = cost(S₅) / (S₅ – Output) = 15 / (5 – 0)

本次迭代中，S₁ 的成本效益最低，因此，添加到输出集的子集 Output = {S₁}，其元素为 {1, 2, 3, 4}。

步骤 2

找到输出集中新元素的每个集合的成本效益，

S₂ = cost(S₂) / (S₂ – Output) = 10 / (5 – 4)
S₃ = cost(S₃) / (S₃ – Output) = 20 / (7 – 4)
S₄ = cost(S₄) / (S₄ – Output) = 12 / (5 – 4)
S₅ = cost(S₅) / (S₅ – Output) = 15 / (5 – 4)

本次迭代中，S₃ 的成本效益最低，因此，添加到输出集的子集 Output = {S₁, S₃}，其元素为 {1, 2, 3, 4, 5, 7, 9, 11, 13}。

步骤 3

找到输出集中新元素的每个集合的成本效益，

S₂ = cost(S₂) / (S₂ – Output) = 10 / |(5 – 9)|
S₄ = cost(S₄) / (S₄ – Output) = 12 / |(5 – 9)|
S₅ = cost(S₅) / (S₅ – Output) = 15 / |(5 – 9)|

本次迭代中，S₂ 的成本效益最低，因此，添加到输出集的子集 Output = {S₁, S₃, S₂}，其元素为 {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13}。

步骤 4

找到输出集中新元素的每个集合的成本效益，

S₄ = cost(S₄) / (S₄ – Output) = 12 / |(5 – 11)|
S₅ = cost(S₅) / (S₅ – Output) = 15 / |(5 – 11)|

本次迭代中，S₄ 的成本效益最低，因此，添加到输出集的子集 Output = {S₁, S₃, S₂, S₄}，其元素为 {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 16, 20}。

步骤 5

找到输出集中新元素的每个集合的成本效益，

S₅ = cost(S₅) / (S₅ – Output) = 15 / |(5 – 14)|

本次迭代中，S₅ 的成本效益最低，因此，添加到输出集的子集 Output = {S₁, S₃, S₂, S₄, S₅}，其元素为 {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 16, 20}。

最终覆盖有限全集中的所有元素的输出为 Output = {S₁, S₃, S₂, S₄, S₅}。

实现

以下是上述方法在各种编程语言中的实现：

C C++ Java Python

#include <stdio.h>
#define MAX_SETS 100
#define MAX_ELEMENTS 1000
int setCover(int X[], int S[][MAX_ELEMENTS], int numSets, int numElements, int output[]) {
   int U[MAX_ELEMENTS];
   for (int i = 0; i < numElements; i++) {
      U[i] = X[i];
   }
   int selectedSets[MAX_SETS];
   for (int i = 0; i < MAX_SETS; i++) {
      selectedSets[i] = 0; // Initialize all to 0 (not selected)
   }
   int outputIdx = 0;
   while (outputIdx < numSets) {  // Ensure we don't exceed the maximum number of sets
      int maxIntersectionSize = 0;
      int selectedSetIdx = -1;
      // Find the set Si with the maximum intersection with U
      for (int i = 0; i < numSets; i++) {
         if (selectedSets[i] == 0) { // Check if the set is not already selected
            int intersectionSize = 0;
            for (int j = 0; j < numElements; j++) {
               if (U[j] && S[i][j]) {
                  intersectionSize++;
               }
            }
            if (intersectionSize > maxIntersectionSize) {
               maxIntersectionSize = intersectionSize;
               selectedSetIdx = i;
            }
         }
      }
      // If no set found, break from the loop
      if (selectedSetIdx == -1) {
          break;
      }
      // Mark the selected set as "selected" in the array
      selectedSets[selectedSetIdx] = 1;
      // Remove the elements covered by the selected set from U
      for (int j = 0; j < numElements; j++) {
          U[j] = U[j] - S[selectedSetIdx][j];
      }
      // Add the selected set to the output
      output[outputIdx++] = selectedSetIdx;
   }
   return outputIdx;
}
int main() {
   int X[MAX_ELEMENTS] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
   int S[MAX_SETS][MAX_ELEMENTS] = {
      {1, 1, 0, 0, 0, 0, 0, 0, 0, 0},
      {0, 1, 1, 1, 0, 0, 0, 0, 0, 0},
      {0, 0, 0, 1, 1, 1, 0, 0, 0, 0},
      {0, 0, 0, 0, 0, 1, 1, 1, 0, 0},
      {0, 0, 0, 0, 0, 0, 0, 1, 1, 1}
   };
   int numSets = 5;
   int numElements = 10;
   int output[MAX_SETS];
   int numSelectedSets = setCover(X, S, numSets, numElements, output);
   printf("Selected Sets: ");
   for (int i = 0; i < numSelectedSets; i++) {
      printf("%d ", output[i]);
   }
   printf("\n");
   return 0;
}

输出

Selected Sets: 1 2 3 4 0

#include <iostream>
#include <vector>
using namespace std;
#define MAX_SETS 100
#define MAX_ELEMENTS 1000
// Function to find the set cover using the Approximate Greedy Set Cover algorithm
int setCover(int X[], int S[][MAX_ELEMENTS], int numSets, int numElements, int output[])
{
   int U[MAX_ELEMENTS];
   for (int i = 0; i < numElements; i++) {
      U[i] = X[i];
   }
   int selectedSets[MAX_SETS];
   for (int i = 0; i < MAX_SETS; i++) {
      selectedSets[i] = 0; // Initialize all to 0 (not selected)
   }
   int outputIdx = 0;
   while (outputIdx < numSets) {  // Ensure we don't exceed the maximum number of sets
      int maxIntersectionSize = 0;
      int selectedSetIdx = -1;
      // Find the set Si with maximum intersection with U
      for (int i = 0; i < numSets; i++) {
         if (selectedSets[i] == 0) { // Check if the set is not already selected
            int intersectionSize = 0;
            for (int j = 0; j < numElements; j++) {
               if (U[j] && S[i][j]) {
                  intersectionSize++;
               }
            }
            if (intersectionSize > maxIntersectionSize) {
               maxIntersectionSize = intersectionSize;
               selectedSetIdx = i;
            }
         }
      }
      // If no set found, break from the loop
      if (selectedSetIdx == -1) {
         break;
      }
      // Mark the selected set as "selected" in the array
      selectedSets[selectedSetIdx] = 1;
      // Remove the elements covered by the selected set from U
      for (int j = 0; j < numElements; j++) {
         U[j] = U[j] - S[selectedSetIdx][j];
      }
      // Add the selected set to the output
      output[outputIdx++] = selectedSetIdx;
   }
   return outputIdx;
}
int main()
{
   int X[MAX_ELEMENTS] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
   int S[MAX_SETS][MAX_ELEMENTS] = {
      {1, 1, 0, 0, 0, 0, 0, 0, 0, 0},
      {0, 1, 1, 1, 0, 0, 0, 0, 0, 0},
      {0, 0, 0, 1, 1, 1, 0, 0, 0, 0},
      {0, 0, 0, 0, 0, 1, 1, 1, 0, 0},
      {0, 0, 0, 0, 0, 0, 0, 1, 1, 1}
   };
   int numSets = 5;
   int numElements = 10;
   int output[MAX_SETS];
   int numSelectedSets = setCover(X, S, numSets, numElements, output);
   cout << "Selected Sets: ";
   for (int i = 0; i < numSelectedSets; i++) {
       cout << output[i] << " ";
   }
   cout << endl;
   return 0;
}

输出

Selected Sets: 1 2 3 4 0

import java.util.*;
public class SetCover {
   public static List<Integer> setCover(int[] X, int[][] S) {
      Set<Integer> U = new HashSet<>();
      for (int x : X) {
         U.add(x);
      }
      List<Integer> output = new ArrayList<>();
      while (!U.isEmpty()) {
         int maxIntersectionSize = 0;
         int selectedSetIdx = -1;
         for (int i = 0; i < S.length; i++) {
            int intersectionSize = 0;
            for (int j = 0; j < S[i].length; j++) {
               if (U.contains(S[i][j])) {
                  intersectionSize++;
               }
            }
            if (intersectionSize > maxIntersectionSize) {
               maxIntersectionSize = intersectionSize;
               selectedSetIdx = i;
            }
         }
         if (selectedSetIdx == -1) {
            break;
         }
         for (int j = 0; j < S[selectedSetIdx].length; j++) {
            U.remove(S[selectedSetIdx][j]);
         }
         output.add(selectedSetIdx);
      }
      return output;
   }
public static void main(String[] args) {
   int[] X = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
   int[][] S = {
      {1, 2},
      {2, 3, 4},
      {4, 5, 6},
      {6, 7, 8},
      {8, 9, 10}
   };
   List<Integer> selectedSets = setCover(X, S);
   System.out.print("Selected Sets: ");
   for (int idx : selectedSets) {
      System.out.print(idx + " ");
   }
   System.out.println();
   }
}

输出

Selected Sets: 1 3 4 0 2

def set_cover(X, S):
    U = set(X)
    output = []
    while U:
        max_intersection_size = 0
        selected_set_idx = -1
        for i, s in enumerate(S):
            intersection_size = len(U.intersection(s))
            if intersection_size > max_intersection_size:
                max_intersection_size = intersection_size
                selected_set_idx = i
        if selected_set_idx == -1:
            break
        U = U - set(S[selected_set_idx])
        output.append(selected_set_idx)
    return output
if __name__ == "__main__":
    X = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    S = [
        {1, 2},
        {2, 3, 4},
        {4, 5, 6},
        {6, 7, 8},
        {8, 9, 10}
    ]
    selected_sets = set_cover(X, S)
    print("Selected Sets:", selected_sets)

输出

Selected Sets: 1 3 4 0 2

打印页面