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OBST问题的解法是动态规划，用到了3层循环，

第一层循环变量是子树的节点个数 l

第二层循环的变量是子树的起点位置i，i即是子树的左边界，j是子树的右边界

第三层循环的变量是子树的根节点位置r

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 #define INF 9999;
 5
 6 void Optimal_BST(double p[],int n){
 7     int i,j,l,r;
 8     int root[n+1][n+1];
 9     double e[n+2][n+2];
10     double w[n+2][n+2];
11     for(i=1;i<=n+1;i++){
12         e[i][i-1]=0;
13         w[i][i-1]=0;
14     }
15
16     for(l=1;l<=n;l++){
17         for(i=1;   i<=n-l+1;  i++){  // i,j between 1 and n+1，i是左边界，j是右边界
18             j=i+l-1;
19             e[i][j]=INF;
20             w[i][j]=w[i][j-1]+p[j];
21             for(r=i;r<=j;r++){
22                 double temp=e[i][r-1]+e[r+1][j]+w[i][j];
23                 if(temp<e[i][j]){
24                     e[i][j]=temp;
25                     root[i][j]=r;
26                 }
27             }
28         }
29     }
30     printf("\n\ne table\n");
31     for(i=1;i<=n;i++){
32         for(j=1;j<=n;j++){
33             if(i>j)printf("          ");
34             else printf("%f  ",e[i][j]);
35         }
36         printf("\n");
37     }
38
39     printf("\n\nw table\n");
40     for(i=1;i<=n;i++){
41         for(j=1;j<=n;j++){
42             if(i>j)printf("          ");
43             else printf("%f  ",w[i][j]);
44         }
45         printf("\n");
46     }
47
48     printf("\n\nroot table\n");
49     for(i=1;i<=n;i++){
50         for(j=1;j<=n;j++){
51             if(i>j)printf("  ");
52             else printf("%d ",root[i][j]);
53         }
54         printf("\n");
55     }
56 }
57
58 int main(){
59     int n=5;
60     double p[6]={-1,0.25,0.2,0.05,0.2,0.3};
61     Optimal_BST(p,n);
62     return 0;
63 }