I have got a fast way to finish book-store

[Angel-Jia]

This method can pass all tests in less one second.
First, I get all all possible method to buy books using DFS. For example, I have three books [1, 1, 2], then I got three methods to buy them:

1. [1], [1], [2]
2. [1], [1, 2]
3. [1, 2], [1]

Then I calculate prices of those methods and got the answer.
Here is the DFS code:

fn DFS(books: &[u32]){
let mut categories: Vec<u32> = vec![0; 5]; // 5 book series and the number of each
    for book in books.iter(){
        categories[(book - 1) as usize] += 1;
    }
    
let all_subsets: Vec<Vec<Vec<usize>>> = get_all_subsets(&mut categories, books.len());
}

fn get_all_subsets(categories: &mut Vec<u32>, total_books: usize) -> Vec<Vec<Vec<usize>>>{
    let mut out_tmp: Vec<Vec<usize>> = Vec::new();  
    let mut ret: Vec<Vec<Vec<usize>>> = Vec::new(); 
    recursive(categories, &mut out_tmp, &mut ret, total_books, 0);
    ret
}

pub fn recursive(categories: &mut Vec<u32>, out: &mut Vec<Vec<usize>>, ret:&mut Vec<Vec<Vec<usize>>>, total_books: usize, num_books: usize){
    if total_books == num_books{
        ret.push(out.clone());
        return;
    }
    let mut books_num = num_books;
    let mut book_set: Vec<usize> = Vec::new();
    for i in 0..5{
        if categories[i] != 0{
            book_set.push(i);
            categories[i] -= 1;
            books_num += 1;
            out.push(book_set.clone());
            recursive(categories, out, ret, total_books, books_num);
            out.pop();
        }
    }
    for item in book_set.iter(){
        categories[*item] += 1;
    }
}