算法笔记-----最优二叉搜索树

//
// Created by alim on 2017/12/23.
//

#include<iostream>
#include<cmath>   //求绝对值函数需要引入该头文件
using namespace std;
const int M=1000+5;
double c[M][M],w[M][M],p[M],q[M];
int s[M][M];
int n,i,j,k;
void Optimal_BST()
{
    for(i=1;i<=n+1;i++)
    {
        c[i][i-1]=0.0;
        w[i][i-1]=q[i-1];
    }
    for(int t=1;t<=n;t++)//t为关键字的规模
        //从下标为i开始的关键字到下标为j的关键字
        for(i=1;i<=n-t+1;i++)
        {
            j=i+t-1;
            w[i][j]=w[i][j-1]+p[j]+q[j];
            c[i][j]=c[i][i-1]+c[i+1][j];//初始化
            s[i][j]=i;//初始化
            //选取i+1到j之间的某个下标的关键字作为从i到j的根，如果组成的树的期望值当前最小，则k为从i到j的根节点
            for(k=i+1;k<=j;k++)
            {
                double temp=c[i][k-1]+c[k+1][j];
                if(temp<c[i][j]&&fabs(temp-c[i][j])>1E-6)//C++中浮点数因为精度问题不可以直接比较
                {
                    c[i][j]=temp;
                    s[i][j]=k;//k即为从下标i到j的根节点
                }
            }
            c[i][j]+=w[i][j];
        }
}
void Construct_Optimal_BST(int i,int j,bool flag)
{
    if(flag==0)
    {
        cout<<"S"<<s[i][j]<<" 是根"<<endl;
        flag=1;
    }
    int k=s[i][j];
    //如果左子树是叶子
    if(k-1<i)
    {
        cout<<"e"<<k-1<<" is the left child of "<<"S"<<k<<endl;
    }
        //如果左子树不是叶子
    else
    {
        cout<<"S"<<s[i][k-1]<<" is the left child of "<<"S"<<k<<endl;
        Construct_Optimal_BST(i,k-1,1);
    }
    //如果右子树是叶子
    if(k>=j)
    {
        cout<<"e"<<j<<" is the right child of "<<"S"<<k<<endl;
    }
        //如果右子树不是叶子
    else
    {
        cout<<"S"<<s[k+1][j]<<" is the right child of "<<"S"<<k<<endl;
        Construct_Optimal_BST(k+1,j,1);
    }
}
int main()
{
    cout << "请输入关键字的个数 n:";
    cin >> n;
    cout<<"请依次输入每个关键字的搜索概率:";
    for (i=1; i<=n; i++ )
        cin>>p[i];
    cout << "请依次输入每个虚结点的搜索概率:";
    for (i=0; i<=n; i++)
        cin>>q[i];
    Optimal_BST();
    // /*用于测试
    for(i=1; i<=n+1;i++)
    {
        for (j=1; j<i;j++)
            cout <<"\t" ;
        for(j=i-1;j<=n;j++)
            cout << w[i][j]<<"\t" ;
        cout << endl;
    }
    for(i=1; i<=n+1;i++)
    {
        for (j=1; j<i;j++)
            cout <<"\t" ;
        for(j=i-1; j<=n;j++)
            cout << c[i][j]<<"\t" ;
        cout << endl;
    }
    for(i=1; i<=n;i++)
    {
        for (j=1; j<i;j++)
            cout << "\t" ;
        for(j=i-1; j<=n;j++)
            cout << s[i][j]<<"\t" ;
        cout << endl;
    }
    cout << endl;
    // */用于测试
    cout<<"最小的搜索成本为:"<<c[1][n]<<endl;
    cout<<"最优二叉搜索树为:";
    Construct_Optimal_BST(1,n,0);
    return 0;
}
/**
 * 0.04 0.09 0.08 0.02 0.12 0.14
 * 0.06 0.08 0.10 0.07 0.05 0.05 0.10
 */
/*
 * 请输入关键字的个数 n:6
6
请依次输入每个关键字的搜索概率:0.04 0.09 0.08 0.02 0.12 0.14
0.04 0.09 0.08 0.02 0.12 0.14
请依次输入每个虚结点的搜索概率:0.06 0.08 0.10 0.07 0.05 0.05 0.10
0.06 0.08 0.10 0.07 0.05 0.05 0.10
0.06    0.18    0.37    0.52    0.59    0.76    1
        0.08    0.27    0.42    0.49    0.66    0.9
                0.1     0.25    0.32    0.49    0.73
                        0.07    0.14    0.31    0.55
                                0.05    0.22    0.46
                                        0.05    0.29
                                                0.1
0       0.18    0.55    0.95    1.23    1.76    2.52
        0       0.27    0.67    0.9     1.38    2.09
                0       0.25    0.46    0.94    1.48
                        0       0.14    0.45    0.98
                                0       0.22    0.68
                                        0       0.29
                                                0
0       1       2       2       2       3       5
        0       2       2       3       3       5
                0       3       3       3       5
                        0       4       5       5
                                0       5       6
                                        0       6

最小的搜索成本为:2.52
最优二叉搜索树为:S5 是根
S2 is the left child of S5
S1 is the left child of S2
e0 is the left child of S1
e1 is the right child of S1
S3 is the right child of S2
e2 is the left child of S3
S4 is the right child of S3
e3 is the left child of S4
e4 is the right child of S4
S6 is the right child of S5
e5 is the left child of S6
e6 is the right child of S6

 *
 */