#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]表示最优树T(i,j)的根节点元素的下标
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;
}
}

运行结果如图
以上算法用s[i][j]保存最优子树T(i,j)的根节点。当s[1][n]=k 时,Xk 为最优搜索树的根节点。T(1,n)的子问题为T(1,k-1)和T(k+1,n),同理可知T(1,k-1)的根节点下标为s[1][k-1]。以此类推可得此问题最优搜索树如图:
算法设计课作业,参考《算法设计与分析》,王晓东著。
如有错误,欢迎评论区批评指正。
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