(p231)15.5-4最优二叉搜索树（n^2和n^3）

对于长度为l的所有子数组的root[i+1][j]-root[i][j-1]+1的和其实就是root[n-l+1]
-root[1][l]+n-l=o(n),所以两层循环就是n^2，但是并不知道Knuth的结论是怎么来的，希望知道的分享一下^-^

/*attention if max set to 1000 will cause segmentation fault*/
#include<stdio.h>
#include<stdlib.h>
#define max 100
#define inf 1e20
struct node{
int n;
struct node *l,*r;
};
float optimal_bst(float p[],float q[],float e[][max+1],int root[][max+1],int n)
{
int i,j,r,l;
float w[max+2][max+1];
for (i=1;i<=n+1;i++)
{
e[i][i-1]=q[i-1];
w[i][i-1]=q[i-1];
}
for (l=0;l<=n-1;l++)
for (i=1;i<=n-l;i++)
{
j=i+l;
w[i][j]=w[i][j-1]+p[j]+q[j];
e[i][j]=inf;
for (r=i;r<=j;r++)
if (e[i][r-1]+e[r+1][j]+w[i][j]<e[i][j])
{
e[i][j]=e[i][r-1]+e[r+1][j]+w[i][j];
root[i][j]=r;
}
}
return e[1]
;
}
float optimal_bst2(float p[],float q[],float e[][max+1],int root[][max+1],int n)
{
int i,j,r,l;
float w[max+2][max+1];
for (i=1;i<=n+1;i++)
{
e[i][i-1]=q[i-1];
w[i][i-1]=q[i-1];
}
for (i=1;i<=n;i++)
{
root[i][i]=i;
w[i][i]=w[i][i-1]+p[i]+q[i];
e[i][i]=e[i][i-1]+e[i+1][i]+w[i][i];
}
for (l=1;l<=n-1;l++)
for (i=1;i<=n-l;i++)
{
j=i+l;
w[i][j]=w[i][j-1]+p[j]+q[j];
e[i][j]=inf;
for (r=root[i][j-1];r<=root[i+1][j];r++)
if (e[i][r-1]+e[r+1][j]+w[i][j]<e[i][j])
{
e[i][j]=e[i][r-1]+e[r+1][j]+w[i][j];
root[i][j]=r;
}
}
return e[1]
;
}
int construct_optimal_bst(int root[][max+1],struct node *troot,int l,int r)
{
struct node *new;
int tmp;
new=(struct node *)malloc(sizeof(struct node));
if (l>r)
{
troot->n=r;
troot->l=NULL;
troot->r=NULL;
return 0;
}
tmp=root[l][r];
troot->n=tmp;
troot->l=(struct node *)malloc(sizeof(struct node));
troot->r=(struct node *)malloc(sizeof(struct node));
construct_optimal_bst(root,troot->l,l,tmp-1);
construct_optimal_bst(root,troot->r,tmp+1,r);
return 0;
}
int print(struct node *troot)
{
if (troot==NULL)
{
printf("# ");
return 0;
}
if ((troot->l==NULL)&&(troot->r==NULL))
printf("d%d ",troot->n);
else
printf("k%d ",troot->n);
print(troot->l);
print(troot->r);
return 0;
}
int main(void)
{
float p[max+1]={0,0.15,0.10,0.05,0.10,0.20};
float q[max+1]={0.05,0.10,0.05,0.05,0.05,0.10};
float e[max+2][max+1];
float ans;
int root[max+2][max+1],n;//can be reduced to [max+1][max+1]
n=5;
ans=optimal_bst2(p,q,e,root,n);
printf("%f\n",ans);
struct node *troot;
troot=(struct node *)malloc(sizeof(struct node));
construct_optimal_bst(root,troot,1,n);
print(troot);
printf("\n");
return 0;
}