动态规划失败的测试用例

Question

问题定义：

计算一组中最长的子集对，总和为 10。一旦确定了一对，这两个数字就不能用于形成其他对。输出应该是一个数字，表示此类对中涉及的数字计数。

输入：1 {1,2,8,9,1,9,1,9} 答案：6（3 对，6 个数字，索引 = {0,7},{3,6},{4, 5})

输入：2 {5,5,5,5,5,5,5,5} 答案：8（4 对，8 个数字，索引 = {0,7},{1,6},{2, 5},{3,4})

输入：3 {2,4,3,7,3,8,6,7} 答案：4（2 对，4 个数字，索引 = {0,5},{3,4} 或 {0, 5},{2,3} 或 {2,7},{3,4} 或 {1,6},{2,3} 或 {1,6},{3,4})

我写了以下代码：

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

#define ARRAYSIZE 10
#define SUM 10

int d[ARRAYSIZE][ARRAYSIZE];
int count = 0;
int final[ARRAYSIZE/2];
int reset = 0;
int subset_to_sum(int a[], int m, int n, int sum)
{
    //printf("%d %d %d \t",m,n,d[m][n]);
    if (m >= n)
        return 0;
    if (d[m][n] != -1)
        return d[m][n];

    if (a[m]+a[n] == sum)
    {
        d[m][n] = 1;
        count = count+1;
        //printf("%d\n",count);
        if (m+1>=n-1)
        {
            final[reset] = count;
            reset = reset+1;
            count = 0;
        }
        else
            final[reset] = count;
        return 1+subset_to_sum(a, m+1, n-1, sum);
    }
    else if (a[m] > sum)
        return subset_to_sum(a, m+1, n, sum);
    else if(a[n] > sum)
        return subset_to_sum(a, m, n-1, sum);
    else
        return (subset_to_sum(a, m+1, n, sum)+subset_to_sum(a, m, n-1, sum)+subset_to_sum(a, m+1, n-1, sum));       
}

int main(void)
{
    int i = 0;
    int j;
    int inc;

    //int set[] = {1,2,8,9,1,9,1,9};
    //int set[] = {1,2,8,4,5,9,1,9};
    //int set[] = {2,4,3,7,3,8,6,7};
    //int set[] = {5,5,5,5,5,5,5,5};
    //int set[] = {25,35,45,5,6,7,8,9};
    //int set[] = {1,2,3,4,5,6,7,8,9,10};

    for(i=0;i<ARRAYSIZE;i++)
    {
        for(j=0;j<ARRAYSIZE;j++)
        {
            d[i][j] = -1;
        }
    }
            printf("\n");
    //printf("Total Pairs making sum %d is : %d\n",SUM,subset_to_sum(set, 0, ARRAYSIZE-1, SUM)*2);
    subset_to_sum(set, 0, ARRAYSIZE-1, SUM);

    int max = 0;
    for(i=0;i<4;i++)
    {
        if(final[i]>max)
            max = final[i];
    }
    printf("Subset Pairs making sum %d is: %d\n",SUM,max*2);
    return 0;
}

我的代码输入失败：{1,2,3,4,5,6,7,8,9,10}。有人能指出我方法中的缺陷吗？

谢谢！

Answer 1

我稍微修改了你的代码，你可以试试下面：

int subset_to_sum(int a[], int m, int n, int sum)
{

if (m >= n)
    return 0;

if (a[m]+a[n] == sum)
  return 1+subset_to_sum(a, m+1, n, sum) + subset_to_sum(a, m, n-1, sum);
else
    return subset_to_sum(a, m+1, n, sum) + subset_to_sum(a, m, n-1, sum);       
}