最优二叉树搜索

// zuiyousousuoerchashu.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include<iostream>
using namespace std;
double p[] = { 0,0.15,0.1,0.05,0.10,0.2 };
double q[] = { 0.05,0.10,0.05,0.05,0.05,0.10};
double e[7][7], w[7][7];
int root[7][7];

void option(int n)                    //求最优搜索二叉树
{
	for (int i = 1; i <= n + 1; i++) {
		e[i][i - 1] = q[i - 1];       //j=i-1，则节点左边为空，此时为伪关键字d(i-1)
		w[i][i - 1] = q[i - 1];      //w为搜索代价增量
	}
	for (int l = 1; l <= n; l++) {     //自底向上地计算间隔为1,2,3..个节点时的最优父节点，l为间隔量
		for (int i = 1; i <= n - l + 1; i++) {           //i为起点
			int j = i + l - 1;                          //j为每一段的终点
			e[i][j] = 65535;                            //先设置为一个较大的树
			w[i][j] = w[i][j - 1] + p[j] + q[j];        //利用递归式计算w的值
			for (int r = i; r <= j; r++) {              //此循环从[i,j]中找出使代价最小的最优节点作为父节点，w[i][j]已由上步算出
				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;
				}
			}
		}
	}
}

void construct(int i, int j)  //重建最优二叉树，先递归左子树的节点，再递归右子树的节点
{
	if (i <= j)
	{
		int r = root[i][j];
		cout << r <<endl;
		construct(i, r - 1);
		construct(r+1,j);
	}
}

void construct2(int i, int j)  //重建最优二叉树,分三种情况
{
	if(i==1&&j==5)
		cout << "k" << root[1][5] << "是根" << endl;       
	if (i < j)
	{
		int r = root[i][j];
		if (i != r)
			cout << "k" << root[i][r - 1] << "是k" << r << "的左孩子"<<endl;
		construct2(i, r - 1);                       //先递归左子树
		if (j != r)
			cout << "k" << root[r + 1][j] << "是k" << r << "的右孩子" << endl;
		construct2(r+1,j);                          //后递归右子树
	}
	if (i == j)                                    //i=j时，此时该节点有两个伪关键字作为子节点
	{
		cout<<"d"<<i-1<<"是k"<<i<< "的左孩子" << endl;
		cout << "d" << i<< "是k" << i << "的右孩子" << endl;
	}
	if (i > j)                                   //i>j时，该节点右边没有关键字，因此右边有一个伪关键字子节点
	{
		cout << "d" << j << "是k" << j << "的右孩子" << endl;
	}
}

int main()
{
	option(5);
	//construct(1, 5);
	construct2(1, 5);
	while (1);
    return 0;
}