1) Problem Description
Using the optimal binary search tree to achieve the minimum cost of the search tree. Spend all nodes of the tree each node has a probability value to be searched pi, spend searching a node is pi * (depth (ki) +1), how to construct a binary search tree so that the tree search the minimum is the optimum binary search tree problem. The problem with the idea of dynamic programming can be realized.
Formal definitions: Given n different keywords sorted sequence K = (k1, k2, ..., kn) and therefore (k1 <k2 <k3 <... <kn), we wish to construct a binary search with these keywords. tree. For each keyword ki, we have a probability of pi represents the probability of starting the search. Some of the value of K may no longer be searched, so we also need n + 1 pseudo keyword (d0, d1, d2, ... dn), for each pseudo-keyword search has a probability qi represents the corresponding probability.
example:
The binary search tree has five vertices (x1, x2, x3, x4 , x5).
[X1 <x2 <x3 <x4 < x5 ]
2) The basic idea
3) code implementation
public class example {
public static void main(String[] args) {
double[] p={0,0.15,0.1,0.05,0.1,0.2}; //n=5关键字有5个
double[] q={0.05,0.1,0.05,0.05,0.05,0.1}; //叶子结点有n+1 = 6个
///这里的关键字长度为5
int n = p.length;
System.out.println("输出根节点辅助表");
int[][] root = Optimal_BST(p,q,n-1);
int temp = root.length-1;
for(int i=1;i<temp;i++){
for(int j=1;j<temp;j++){
System.out.print(root[i][j]+"-");
}
System.out.println();
}
printOptimalBST(root,1,5,root[1][5]);
}
/**
* DP在计算最优二叉树的辅助表的算法实现
* @param p
* @param q
* @param n
* @return
*/
private static int[][] Optimal_BST(double[] p, double[] q, int n) {
double[][] e = new double[n+2][n+2];//
double[][] w = new double[n+2][n+2];
int[][] root = new int[n+2][n+2];
//初始化叶子结点的值
for(int i=1;i<=n+1;i++){
e[i][i-1]=q[i-1];
w[i][i-1]=q[i-1];
}
for(int l=1 ; l<=n ; l++){///最外层循环是逐渐的将关键字个数从一个扩展到n个
for(int i=1;i<=n-l+1;i++){
int j=i+l-1;
e[i][j]=Double.MAX_VALUE;
w[i][j]=w[i][j-1]+p[j]+q[j];
for(int r=i;r<=j;r++){
double t = e[i][r-1]+e[r+1][j]+w[i][j];
if(t<e[i][j]){
e[i][j]=t;
root[i][j]=r;///存储根节点的位置
}
}
}
}
System.out.println("输出当前的最小代价："+e[1][n]);
return root;
}
/**
* 构建最优二叉搜索树
* @param root
* @param i
* @param j
* @param k
*/
private static void printOptimalBST(int[][] root, int i, int j, int r) {
int rootChild = root[i][j];
if(rootChild==r){
System.out.println("K"+rootChild+"是根");
printOptimalBST(root,i,rootChild - 1,rootChild);
printOptimalBST(root,rootChild + 1,j,rootChild);
return;
}
if (j < i - 1)
{
return;
}
else if (j == i - 1)//遇到虚拟键
{
if (j < r)
{
System.out.println( "d" + j + "是" + "k" + r + "的左孩子" );
}
else {//j>=r
System.out.println( "d" + j + "是" + "k" + r + "的右孩子" );
}
return;
}
else//遇到内部结点
{
if (rootChild < r)
{
System.out.println ("k" + rootChild + "是" + "k" + r + "的左孩子" );
}
else{
System.out.println ("k" + rootChild + "是" + "k" + r + "的右孩子" );
}
}
printOptimalBST(root,i,rootChild - 1,rootChild);
printOptimalBST(root,rootChild + 1,j,rootChild);
}
}输出根节点辅助表
输出当前的最小代价：2.75
1-1-2-2-2-
0-2-2-2-4-
0-0-3-4-5-
0-0-0-4-5-
0-0-0-0-5-
K2是根
k1是k2的左孩子
d0是k1的左孩子
d1是k1的右孩子
k5是k2的右孩子
k4是k5的左孩子
k3是k4的左孩子
d2是k3的左孩子
d3是k3的右孩子
d4是k4的右孩子
d5是k5的右孩子4) time complexity and space complexity
Time complexity of
O (n- . 3)
spatial complexity of
O (n- 2)
