#include<iostream>
using namespace std;
void OptimalBST(float*p_Node,float*p_NonNode, float m [6][6],float w[6][6],float s[6][6],int NodeNum)
{
    ///初始化
    for(int i = 0;i<=NodeNum;i++){
        w[i+1][i] = p_NonNode[i];
        m[i+1][i] = 0;
    }
    for(int r = 0;r<NodeNum;r++)
        for(int i = 1;i<=NodeNum-r;i++){
            int j = i+r;
            w[i][j] = w[i][j-1]+p_NonNode[j]+p_Node[j];
            m[i][j] = m[i+1][j];
            ///s[i][j]表示最優樹T(i,j)的根節點元素的下標
            s[i][j] = i;
            for(int k = i+1;k<=j;k++){
                float t = m[i][k-1]+m[k+1][j];
                if(t<m[i][j]){
                    //cout<<"m[i][j]="<<m[i][j];
                    //cout<<"    t=" <<t;
                    //cout<<"    k="<<k<<endl;
                    m[i][j] = t;
                    s[i][j] = k;
                }
            }
            m[i][j] += w[i][j];
    }
}
int main()
{
    int NodeNum = 5;
    float p_Node[6] = {0,0.1,0.3,0.1,0.2,0.1};
    float p_NonNode[6] = {0.04,0.02,0.02,0.05,0.06,0.01};
    float m[6][6];
    float w[6][6];
    float s[6][6] = {-1};
    OptimalBST(p_Node,p_NonNode,m,w,s,NodeNum);
    for(int i = 0;i<=NodeNum;i++){
        for(int j = 0;j<=NodeNum;j++){
            cout<<s[i][j]<<" ";
        }
        cout<<endl;
    }
}

 

執行結果如圖

以上演算法用s[i][j]儲存最優子樹T(i,j)的根節點。當s[1][n]=k 時，Xk 為最優搜尋樹的根節點。T(1,n)的子問題為T(1,k-1)和T(k+1,n),同理可知T(1,k-1)的根節點下標為s[1][k-1]。以此類推可得此問題最優搜尋樹如圖：

演算法設計課作業，參考《演算法設計與分析》，王曉東著。

如有錯誤，歡迎評論區批評指正。

轉載請註明出處。