优化子集和实现

Question

我正在使用以下代码解决子集和问题的变体。该问题需要从更大的集合（超集）中生成 11 个整数的子集，并检查它是否与特定值匹配（endsum）。

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

int endsum = 0, supersetsize = 0, done = 0;
int superset[] = {1,30,10,7,11,27,3,5,6,50,45,32,25,67,13,37,19,52,18,9};
int combo = 0;

int searchForPlayerInArray(int arr[], int player) {
    for (int i=0; i<11; i++) {
        if (arr[i] == player) {
            return 1;
        }
    }
    return 0;
}

int sumOfArray(int arr[]) {
    int res = 0;
    for (int i=0; i<11; i++) {
        res+=arr[i];
    }
    return res;
}

void printArray(int arr[], int arrSize) {
    for (int j=0; j<arrSize; j++) {
        printf("%2d ",arr[j]);
    }
    printf("= %d\n",endsum);
}

void permute(int subset[], int pos, int sspos) {
    if (done) { //when a correct solution has been found, stop recursion
        return;
    }
    if (sspos == supersetsize) { // out of possible additions
        return;
    }
    if (pos == 11) { //is the current subset 11 ints long?
        int res = sumOfArray(subset);
        combo++;
        if (res == endsum) { //if the sum of the array matches the wanted sum, print
            printArray(subset,11);
            done = 1;
        }
        return;
    }
    for (int i=sspos; i<supersetsize; i++) {
        //assert(pos < 11);
        //assert(i+1 <= supersetsize);
        subset[pos] = superset[i];
        permute(subset,pos+1,i+1);
    }
}

int main(void) {
    endsum = 110;
    supersetsize = 20;
    int *arr;
    arr = malloc(supersetsize*sizeof(int));
    int i;
    for (i=0; i<supersetsize; i++) {
        arr[i] = 0;
    }

    permute(arr,0,0);

    printf("Combinations: %d",combo);
    return 0;
}

虽然此解决方案适用于小型超集 (

编辑：根据大众需求添加的完整源代码。

Answer 1

仅生成唯一子集的一种方法是按顺序添加超集中的元素，并使用permute 的附加参数（例如supersetPos）来指示您在超集中的位置。这会生成唯一的排序排列。

编辑：AFAIK 在您的示例上正确运行的代码：

#include <stdio.h>

int superset[] = {
 1, 30, 10, 7, 11,
 27, 3, 5, 6, 50,
 45, 32, 25, 67, 13,
 37, 19, 52, 18, 9
};
int supersetsize = 20;
int endsum = 110;
int done = 0;

int sumOfArray(int array[]) {
  int sum = 0;
  for(int i = 0; i < 11; i++)
      sum += array[i];
  return sum;
}

void permute(int subset[], int pos, int sspos) {

    if (pos == 11) { //is the current subset 11 ints long?
        if (sumOfArray(subset) == endsum) { //if the sum of the array matches the wanted sum, print
            for (int j=0; j<11; j++) {
                printf("%d ",subset[j]);
            }
            printf("\n");
            done = 1;
        }
        return;
    }
    for (int i=sspos; i<supersetsize; i++) {
        subset[pos] = superset[i];
        permute(subset,pos+1,i+1);
        if (done) { //when a correct solution has been found, stop recursion
            return;
        }
    }

}
int main() {
  int subset[11] = {0};
  permute(subset, 0, 0);
}

Answer 2

我认为没有办法在比指数时间更好的时间内生成唯一子集。

要有效地解决子集和，您需要使用动态规划。有一些用于子集和的伪多项式时间算法以这种方式工作。这个Wikipedia article 可能会有所帮助。

Answer 3

你可以试试我的代码（我试着只给出一个伪代码而不是完全解决你的作业）：

// array is all the numbers you are looking from them
// length is the number of arrays
// pos is the position of the slot you are going to fill
// max is nomber of slots you have to fill (in your case since you are going for the 11 sets you have to set this value as 11
// sum is the sum of all the values selected until now
// searchbegin is the first element you can pick from your array (I'm using this variable to only generate subarrays of the superset (array))
// target is the targetvalue you are looking for.

void generate_all(int []array, int length, int pos,int max, int sum,int searchbegin,int target)
{
    if max = pos
        if sum = target
            printselectedresults();
    for i:searchbegin->length-max+pos
        if (sum + array[i] < target)
        {
            addtoresults(i);
            generate_all(array,length,pos+1,max,sum+array[i],i+1,target);
            removefromresults(i);
        }
}

有了所有这些信息，我认为您可以轻松地将该代码实现为您的目标语言并使用它。

在我的函数中，所有生成的排列都是超集的子数组，因此不能生成两次排列，而且每个排列至少生成一次。