最优二叉搜索树错误输出

Question

我正在研究一个问题，通过尽可能最好的方式来计算二叉搜索树中所需的最少遍历次数。确实在网上看到了一个解决方案，我明白了，但是我手工对样本输入进行了一些计算，结果没有得到正确的结果。

代码如下

#include <stdio.h>
#include <limits.h>

// A utility function to get sum of array elements freq[i] to freq[j]
int sum(int freq[], int i, int j);

/* A Dynamic Programming based function that calculates minimum cost of
   a Binary Search Tree. */
int optimalSearchTree(int keys[], int freq[], int n)
{
    /* Create an auxiliary 2D matrix to store results of subproblems */
    int cost[n][n];

    /* cost[i][j] = Optimal cost of binary search tree that can be
       formed from keys[i] to keys[j].
       cost[0][n-1] will store the resultant cost */

    // For a single key, cost is equal to frequency of the key
    for (int i = 0; i < n; i++)
        cost[i][i] = freq[i];

    // Now we need to consider chains of length 2, 3, ... .
    // L is chain length.
    for (int L=2; L<=n; L++)
    {
        // i is row number in cost[][]
        for (int i=0; i<=n-L+1; i++)
        {
            // Get column number j from row number i and chain length L
            int j = i+L-1;
            cost[i][j] = INT_MAX;

            // Try making all keys in interval keys[i..j] as root
            for (int r=i; r<=j; r++)
            {
               // c = cost when keys[r] becomes root of this subtree
               int c = ((r > i)? cost[i][r-1]:0) + 
                       ((r < j)? cost[r+1][j]:0) + 
                       sum(freq, i, j);
               if (c < cost[i][j])
                  cost[i][j] = c;
            }
        }
    }
    return cost[0][n-1];
}

// A utility function to get sum of array elements freq[i] to freq[j]
int sum(int freq[], int i, int j)
{
    int s = 0;
    for (int k = i; k <=j; k++)
       s += freq[k];
    return s;
}

// Driver program to test above functions
int main()
{
    int keys[] = {10, 12, 20};
    int freq[] = {34, 8, 50};
    int n = sizeof(keys)/sizeof(keys[0]);
    printf("Cost of Optimal BST is %d ", optimalSearchTree(keys, freq, n));
    return 0;
}

例如对于输入 int 键[] = {1,2,3}; int freq[] = {10,3,1};

I SHOULD GET 18 but i GET 19

对于这个输入 int 键[] = {1,2,3,4}; int freq[] = {5,4,1,200};

I SHOULD GET 225 I GET 226

对于这个输入 int 键[] = {1,2,3,4,5,6}; int freq[] = {33,1,409,2,1,34};

I SHOULD GET 997 but I GET 556

为此输入 int keys[] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}; int freq[] = {5,5,5,5,5,5,5,5,5,5,167,5,5,5,5,5,5,5,5,5};

I Should Get 789 but i GET 532

怎么了？

Answer 1

您对预期的结果有误。以您的第一个示例为例，键 {1, 2, 3} 和频率 {10, 3, 1}；最佳二叉搜索树是：

（括号中给出的频率）。函数 returns 搜索树 sum(frequencies) 次的预期成本，以访问的节点数衡量，它是 (10 * 1) + (3 * 2) + (1 * 3) = 19。每次搜索 key 1 只遍历根节点。每次查找key2都会遍历根节点，在其子节点中找到目标key，一共遍历两个节点。同理，每次搜索key3都会遍历三个节点

根据访问的节点来衡量成本是有意义的，因为必须针对每个访问的节点的键进行比较，但您也可以根据遍历的边来衡量成本。但是，如果您计算进入根的边，则结果是相同的，表示宿主程序指向树的指针。在那种情况下，每个节点访问都涉及遍历该节点的传入边，消除了节点访问或边遍历计算成本之间的任何区别。

使用键 {1,2,3,4} 和频率 {5,4,1,200}，最佳二叉搜索树是

费用为 (200 * 1) + (5 * 2) + (4 * 3) + (1 * 4) = 226。

使用键 {1,2,3,4,5,6} 和频率 {33,1,409,2,1,34}:

            3 (409)
          /     \
         /       \
        1 (33)    6 (34)
         \       /
          2 (1)  4 (2)
                  \
                   5(1)

费用为 (409 * 1) + ((33 + 34) * 2) + ((1 + 2) * 3) + (1 * 4) = 556。

程序 returns 每种情况下的正确结果。

我将 20 键示例作为练习留给您。对于这种情况，有多个最优搜索树，但所有的成本都是 532，正如程序给出的那样。

最优二叉搜索树错误输出

Optimal Binary Search Tree Wrong Output

c

algorithm

dynamic-programming