动态规划：最优二叉搜索树（手写例题附代码）

本篇直接上例题，至于理论大家可以搜索优快云大佬的文章。

参考代码（伪代码）

题目：

 

记录的计算的中间与最终结果：

代码

//最优二叉查找树
 
#include <iostream>
 
using namespace std;
 
const int MaxVal = 9999;
 
const int n = 7;
//搜索到根节点和虚拟键的概率
double p[n + 1] = {-1,0.04,0.06,0.08,0.02,0.10,0.12,0.14};
double q[n + 1] = {0.06,0.06,0.06,0.06,0.05,0.05,0.05,0.05};
 
int root[n + 1][n + 1];//记录根节点
double w[n + 2][n + 2];//子树概率总和
double e[n + 2][n + 2];//子树期望代价
 
void optimalBST(double *p,double *q,int n)
{
	//初始化只包括虚拟键的子树
	for (int i = 1;i <= n + 1;++i)
	{
		w[i][i - 1] = q[i - 1];
		e[i][i - 1] = q[i - 1];
	}
 
	//由下到上，由左到右逐步计算
	for (int len = 1;len <= n;++len)
	{
		for (int i = 1;i <= n - len + 1;++i)
		{
			int j = i + len - 1;
			e[i][j] = MaxVal;
			w[i][j] = w[i][j - 1] + p[j] + q[j];
			//求取最小代价的子树的根
			for (int k = i;k <= j;++k)
			{
				double temp = e[i][k - 1] + e[k + 1][j] + w[i][j];
				if (temp < e[i][j])
				{
					e[i][j] = temp;
					root[i][j] = k;
				}
			}
		}
	}
}
 
//输出最优二叉查找树所有子树的根
void printRoot()
{
	cout << "各子树的根：" << endl;
	for (int i = 1;i <= n;++i)
	{
		for (int j = 1;j <= n;++j)
		{
			cout << root[i][j] << " ";
		}
		cout << endl;
	}
	cout << endl;
}
 
//打印最优二叉查找树的结构
//打印出[i,j]子树，它是根r的左子树和右子树
void printOptimalBST(int i,int j,int r)
{
	int rootChild = root[i][j];//子树根节点
	if (rootChild == root[1][n])
	{
		//输出整棵树的根
		cout << "k" << rootChild << "是根" << endl;
		printOptimalBST(i,rootChild - 1,rootChild);
		printOptimalBST(rootChild + 1,j,rootChild);
		return;
	}
 
	if (j < i - 1)
	{
		return;
	}
	else if (j == i - 1)//遇到虚拟键
	{
		if (j < r)
		{
			cout << "d" << j << "是" << "k" << r << "的左孩子" << endl;
		}
		else
			cout << "d" << j << "是" << "k" << r << "的右孩子" << endl;
		return;
	}
	else//遇到内部结点
	{
		if (rootChild < r)
		{
			cout << "k" << rootChild << "是" << "k" << r << "的左孩子" << endl;
		}
		else
			cout << "k" << rootChild << "是" << "k" << r << "的右孩子" << endl;
	}
 
	printOptimalBST(i,rootChild - 1,rootChild);
	printOptimalBST(rootChild + 1,j,rootChild);
}
 
int main()
{
	optimalBST(p,q,n);
	printRoot();
	cout << "最优二叉树结构：" << endl;
	printOptimalBST(1,n,-1);
}