算法java实现--动态规划--最优二叉树搜索

最优二叉搜索树算法的java实现（动态规划法）

具体问题描述以及C/C++实现参见网址

http://blog.csdn.net/liufeng_king/article/details/8694652

未改进版的java代码如下：

/**
 * 最优二叉搜索树（动态规划）---未改进版
 * @author Lican
 *
 */
public class BinarySearchTree {
	public void optimalBinarySearchTree(float[] a,float[] b,float[][] m,int[][] s,float[][] w){
		int n=a.length-1;		
		for(int i=0;i<=n;i++){//初始化构造无内部节点的情况
			w[i+1][i]=a[i];
			m[i+1][i]=0;
		}
		for(int r=0;r<n;r++){//r为i与j之间的差值
			for(int i=1;i<=n-r;i++){
				int j=i+r;
				//i，j之间距离为r时，首选i为根，其左子树为空，右子树为节点
				w[i][j]=w[i][j-1]+a[j]+b[j];//计算w[i][j]
				m[i][j]=m[i+1][j];
				s[i][j]=i;
				for(int k=i+1;k<=j;k++){//i<=k<=j,通过k循环，找到min{m(i,k-1)+m(k+1,j)}的值
					float temp=m[i][k-1]+m[k+1][j];
					if(temp<m[i][j]){
						m[i][j]=temp;
						s[i][j]=k;//k作为根节点
					}
				}
				m[i][j]+=w[i][j];//m(i,j)=wi,j+min{m(i,k-1)+m(k+1,j)}
			}
		}
	}
	public void backtrace(int[][] s,int n,int i,int j,int p,String str){
		int k=s[i][j];
		if(k>0){
			if(p==0){
				System.out.println("root is:"+k);
			}else{
				System.out.println(str+" of "+p+" is "+k+"; and (i:j) is "+i+":"+j);
			}
			int t=k-1;
			if(t>=i&&t<=n)
				backtrace(s,n,i,t,k,"left");
			t=k+1;
			if(t<=j)
				backtrace(s,n,t,j,k,"right");
		}
	}
	public static void main(String[] args) {
		float a[] = {0.15f,0.1f,0.05f,0.05f};//a,b的下标都是从0开始
		float b[] = {0.00f,0.5f,0.1f,0.05f};
		int n=a.length-1;
		float[][] m=new float[n+2][n+2];
		int[][] s=new int[n+2][n+2];//因为w[i+1][i]存在，i最大为n，即可以从0.....n,n+1,共n+2个
		float[][] w=new float[n+2][n+2];
		BinarySearchTree bi = new BinarySearchTree();
		bi.optimalBinarySearchTree(a, b, m, s, w);
		System.out.println("二叉搜索树最小平均路长为："+m[1][n]);
		bi.backtrace(s, n, 1, n, 0, "0");
	}

}

算法实现结果如图：

===================================================================================================================================

改进后的java代码如下（具体改进的地方，参见引文中的介绍）

/**
 * 最优二叉搜索树（动态规划）---改进版
 * @author Lican
 *
 */
public class BinarySearchTree2 {
	public void optimalBinarySearchTree(float[] a,float[] b,float[][] m,int[][] s,float[][] w){
		int n=a.length-1;		
		for(int i=0;i<=n;i++){//初始化构造无内部节点的情况
			w[i+1][i]=a[i];
			m[i+1][i]=0;
			s[i+1][i]=0;
		}
		for(int r=0;r<n;r++){//r为i与j之间的差值
			for(int i=1;i<=n-r;i++){
				int j=i+r;
				/*以下为 改进后的地方*/
				int i1=s[i][j-1]>i?s[i][j-1]:i;
				int j1=s[i+1][j]>i?s[i+1][j]:j;
				//i，j之间距离为r时，首选i为根，其左子树为空，右子树为节点
				w[i][j]=w[i][j-1]+a[j]+b[j];//计算w[i][j]
				m[i][j]=m[i][i1-1]+m[i1+1][j];//k=i1时的情况
				s[i][j]=i1;
				 //左子树为节点：i1,i1+1……k-1,右子树为节点：k+1,k+2,……j1
				for(int k=i1+1;k<=j1;k++){//i1<=k<=j1,通过k循环，找到min{m(i,k-1)+m(k+1,j)}的值
					float temp=m[i][k-1]+m[k+1][j];
					if(temp<m[i][j]){
						m[i][j]=temp;
						s[i][j]=k;//k作为根节点
					}
				}
				m[i][j]+=w[i][j];//m(i,j)=wi,j+min{m(i,k-1)+m(k+1,j)}
			}
		}
	}
	public void backtrace(int[][] s,int n,int i,int j,int p,String str){
		int k=s[i][j];
		if(k>0){
			if(p==0){
				System.out.println("root is:"+k);
			}else{
				System.out.println(str+" of "+p+" is "+k+"; and (i:j) is "+i+":"+j);
			}
			int t=k-1;
			if(t>=i&&t<=n)
				backtrace(s,n,i,t,k,"left");
			t=k+1;
			if(t<=j)
				backtrace(s,n,t,j,k,"right");
		}
	}
	public static void main(String[] args) {
		float a[] = {0.15f,0.1f,0.05f,0.05f};//a,b的下标都是从0开始
		float b[] = {0.00f,0.5f,0.1f,0.05f};
		int n=a.length-1;
		float[][] m=new float[n+2][n+2];
		int[][] s=new int[n+2][n+2];//因为w[i+1][i]存在，i最大为n，即可以从0.....n,n+1,共n+2个
		float[][] w=new float[n+2][n+2];
		BinarySearchTree bi = new BinarySearchTree();
		bi.optimalBinarySearchTree(a, b, m, s, w);
		System.out.println("二叉搜索树最小平均路长为："+m[1][n]);
		bi.backtrace(s, n, 1, n, 0, "0");
	}

}

运行结果与未改进版相同。