91影视

Consider$\text{CUT}\left(i,j\right)$ be the cut made in order to make maximum selling cost.

Base case is when no cut is made,$i=j=0$ which give $\text{CUT}\left(i,\text{鈥�}j\right)=0$.

To cut horizontally, along$x$, the recursive relation is

$\text{CUT}\left(i,j\right)=\text{max}\left(\text{CUT}\left(i,j\right),\text{CUT}\left(i_{cut},j\right)+\text{CUT}\left(i鈭�i_{cut},j\right)\right)$

To cut vertically, along,$y$ the recursive relation is

$\text{CUT}\left(i\text{,鈥�}j\right)=\text{max}\left(\text{CUT}\left(i\text{,鈥�}j\right),\text{CUT}\left(i\text{,鈥�}{j}_{cut}\right)+\text{CUT}\left(i,j鈭�j_{cut}\right)\right)$

From these two recurrence relation, dimension of cloth required to make maximum selling price.

The algorithm is given as follows:

$\begin{array}{l}\text{for}\left(i=0\text{鈥则辞}x-1\right)\\ \text{鈥勨赌勨赌刦辞谤}\left(j=0\text{鈥则辞鈥�}Y鈭�1\right)\\ \text{鈥勨�勨�勨�勨赌勨赌刦辞谤}\left(i_{cut}=1\text{鈥则辞鈥�}i鈭�1\right)\\ \text{鈥勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌�}CUT\left(i,j\right)=\text{max}\left(CUT\left(i,j\right),CUT\left(i_{cut},j\right)+CUT\left(i鈭�i_{cut},j\right)\right)\end{array}$

$\begin{array}{l}\text{鈥勨�勨�勨�勨赌勨赌刦辞谤}\left(j_{cut}=1\text{鈥则辞}j鈭�1\right)\\ \text{鈥勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌�}CUT\left(i,j\right)=\text{max}\left(CUT\left(i,j\right),CUT\left(i,j_{cut}\right)+CUT\left(i,j鈭�j_{cut}\right)\right)\\ \text{鈥勨�勨�勨�勨赌勨赌刦辞谤}\left(\text{颈迟别尘鈥刬苍滨罢贰惭厂}\right)\\ \text{鈥勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌刬f}\left(ite{m}_{dimension}=\left(i,j\right)\right)\\ \text{鈥勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨赌勨�勨�勨��}CUT\left(i,j\right)=\text{max}\left(CUT\left(i,j\right),c_i\right)\\ \text{谤别迟耻谤苍鈥�}CUT\left(X鈭�1,Y鈭�1\right)\end{array}$

The algorithm returns the desired size of cloth need to earn maximum selling price.

The inner three $for\left(\right)$loops run$X,\text{鈥�}Y$ and$n$ times, so runtime for inner loops are $O\left(X+Y+n\right)$Now outer loops run in nesting, that give $O\left(XY\right)$.

Thus, effective runtime is $O\left(XY\left(X+Y+n\right)\right)$.