Geeks 面试题 ： Optimal Binary Search Tree

Dynamic Programming | Set 24 (Optimal Binary Search Tree)

Given a sorted array keys[0.. n-1] of search keys and an array freq[0.. n-1] of frequency counts, where freq[i] is the number of searches to keys[i]. Construct a binary search tree of all keys such that the total cost of all the searches is as small as possible.

Let us first define the cost of a BST. The cost of a BST node is level of that node multiplied by its frequency. Level of root is 1.

Example 1
Input:  keys[] = {10, 12}, freq[] = {34, 50}
There can be following two possible BSTs 
        10                       12
          \                     / 
           12                 10
          I                     II
Frequency of searches of 10 and 12 are 34 and 50 respectively.
The cost of tree I is 34*1 + 50*2 = 134
The cost of tree II is 50*1 + 34*2 = 118 

Example 2
Input:  keys[] = {10, 12, 20}, freq[] = {34, 8, 50}
There can be following possible BSTs
    10                12                 20         10              20
      \             /    \              /             \            /
      12          10     20           12               20         10  
        \                            /                 /           \
         20                        10                12             12  
     I               II             III             IV             V
Among all possible BSTs, cost of the fifth BST is minimum.  
Cost of the fifth BST is 1*50 + 2*34 + 3*8 = 142

1) Optimal Substructure:
The optimal cost for freq[i..j] can be recursively calculated using following formula.

We need to calculate optCost(0, n-1) to find the result.

The idea of above formula is simple, we one by one try all nodes as root (r varies from i to j in second term). When we make rth node as root, we recursively calculate optimal cost from i to r-1 and r+1 to j.
We add sum of frequencies from i to j (see first term in the above formula), this is added because every search will go through root and one comparison will be done for every search.

http://www.geeksforgeeks.org/dynamic-programming-set-24-optimal-binary-search-tree/

关键：

如何计算整个优化BST的查找次数，这里不用按上面说的一层一层地去计算，

而是巧妙地计算：每增加一个根节点的时候，根节点下面的所有节点的访问次数都需要增加1倍，那么只需要把这些节点的总和加起来就可以了，因为根节点的访问次数也是一倍，所以可以连根节点也加进球，就得到最后的总访问频率了。

经验：一些看起来简单的小函数，要用最优的方法实现就不是那么容易了。

原网站的时间效率是O(n*n*n*n)，原文也提到可以提高到O(n*n*n)。

下面给出两个程序，时间效率都提高到了O(n*n*n)。

第一个程序只需要修改一下计算sum的位置就可以了，

第二个程序增加计算了一个sumTable，记录了所有sum，实际运行效率会稍微快点。

int sum(int freq[], int i, int j)
{
	int s = 0;
	for (; i <= j; i++)
	{
		s +=freq[i];
	}
	return s;
}

int optimalSearchTree(int keys[], int freq[], int n)
{
	vector<vector<int> > tbl(n, vector<int>(n, INT_MAX));
	for (int i = 0; i < n; i++)
	{
		tbl[i][i] = freq[i];
	}

	for (int d = 2; d <= n; d++)
	{
		int i = 0, j = d-1;
		for (int r = 0; r <= n-d; r++, i++, j++)
		{
			int s = sum(freq, i, j);
			for (int k = i; k < d; k++)
			{
				int t = (k > i? tbl[i][k-1]:0)+(k < j? tbl[k+1][j]:0)+s;
				tbl[i][j] = min(tbl[i][j], t);
			}
		}
	}
	return tbl[0][n-1];
}

//优化sum
vector<vector<int> > sumTable(int freq[], int n)
{
	vector<vector<int> > tbl(n, vector<int>(n));
	for (int i = 0; i < n; i++)
	{
		tbl[i][i] = freq[i];
	}
	for (int i = 0; i < n; i++)
	{
		for (int j = i+1; j < n; j++)
		{
			tbl[i][j] = tbl[i][j-1] + freq[j];
		}
	}
	return tbl;
}

int optimalSearchTree_2(int keys[], int freq[], int n)
{
	vector<vector<int> > tbl(n, vector<int>(n, INT_MAX));
	for (int i = 0; i < n; i++)
	{
		tbl[i][i] = freq[i];
	}
	vector<vector<int> > sum_tbl = sumTable(freq, n);
	for (int d = 2; d <= n; d++)
	{
		int i = 0, j = d-1;
		for (int r = 0; r <= n-d; r++, i++, j++)
		{
			for (int k = i; k < d; k++)
			{
				int t = (k > i? tbl[i][k-1]:0)+(k < j? tbl[k+1][j]:0)+sum_tbl[i][j];
				tbl[i][j] = min(tbl[i][j], t);
			}
		}
	}
	return tbl[0][n-1];
}

测试：

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\n", optimalSearchTree(keys, freq, n));

	printf("Cost of Optimal BST is %d\n", optimalSearchTree_2(keys, freq, n));

	system("pause");
	return 0;
}