Luo Gu P1040 plus binary solution to a problem

dp can

\ (F [i] [j] \) represents the i to j plus

Equivalent to the interval dp

#include<cstdio>
using namespace std;
int v[50];
int f[55][55];
int root[55][55];
void print(int l,int r)
{
    if(l>r)return ;
    if(l==r)
    {
        printf("%d ",r);
        return ;
    } 
    int tmp=root[l][r];
    printf("%d ",tmp);
    print(l,tmp-1);
    print(tmp+1,r);
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&v[i]);
        f[i][i]=v[i];
        f[i][i-1]=1;
    }
    for(int i=n-1;i>=1;i--)
    {
        for(int j=i+1;j<=n;j++)
        {
            for(int k=i;k<=j;k++)
            {
                if(f[i][j]<(f[i][k-1]*f[k+1][j]+f[k][k]))//k为根节点
                {
                    f[i][j]=f[i][k-1]*f[k+1][j]+f[k][k];
                    root[i][j]=k;
                }
            }
        }
    }
    printf("%d\n",f[1][n]);
    print(1,n);//输出中序遍历
    return 0; 
}