2312. Selling Pieces of Wood
Description
You are given two integers m and n that represent the height and width of a rectangular piece of wood. You are also given a 2D integer array prices, where prices[i] = [hi, wi, pricei] indicates you can sell a rectangular piece of wood of height hi and width wi for pricei dollars.
To cut a piece of wood, you must make a vertical or horizontal cut across the entire height or width of the piece to split it into two smaller pieces. After cutting a piece of wood into some number of smaller pieces, you can sell pieces according to prices. You may sell multiple pieces of the same shape, and you do not have to sell all the shapes. The grain of the wood makes a difference, so you cannot rotate a piece to swap its height and width.
Return the maximum money you can earn after cutting an m x n piece of wood.
Note that you can cut the piece of wood as many times as you want.
Example 1:
Input: m = 3, n = 5, prices = [[1,4,2],[2,2,7],[2,1,3]] Output: 19 Explanation: The diagram above shows a possible scenario. It consists of: - 2 pieces of wood shaped 2 x 2, selling for a price of 2 * 7 = 14. - 1 piece of wood shaped 2 x 1, selling for a price of 1 * 3 = 3. - 1 piece of wood shaped 1 x 4, selling for a price of 1 * 2 = 2. This obtains a total of 14 + 3 + 2 = 19 money earned. It can be shown that 19 is the maximum amount of money that can be earned.
Example 2:
Input: m = 4, n = 6, prices = [[3,2,10],[1,4,2],[4,1,3]] Output: 32 Explanation: The diagram above shows a possible scenario. It consists of: - 3 pieces of wood shaped 3 x 2, selling for a price of 3 * 10 = 30. - 1 piece of wood shaped 1 x 4, selling for a price of 1 * 2 = 2. This obtains a total of 30 + 2 = 32 money earned. It can be shown that 32 is the maximum amount of money that can be earned. Notice that we cannot rotate the 1 x 4 piece of wood to obtain a 4 x 1 piece of wood.
Constraints:
1 <= m, n <= 2001 <= prices.length <= 2 * 104prices[i].length == 31 <= hi <= m1 <= wi <= n1 <= pricei <= 106- All the shapes of wood
(hi, wi)are pairwise distinct.
Solutions
Solution 1: Memoization Search
First, we define a 2D array $d$, where $d[i][j]$ represents the price of a wood block with height $i$ and width $j$.
Then, we design a function $dfs(h, w)$ to denote the maximum amount of money obtained by cutting a wood block with height $h$ and width $w$. The answer will be $dfs(m, n)$.
The process of function $dfs(h, w)$ is as follows:
If $(h, w)$ has been calculated before, return the answer directly.
Otherwise, initialize the answer as $d[h][w]$, then enumerate the cutting positions, calculate the maximum amount of money obtained by cutting the wood block into two pieces, and take the maximum value.
The time complexity is $O(m \times n \times (m + n) + p)$, and the space complexity is $O(m \times n)$. Here, $p$ represents the length of the price array, while $m$ and $n$ represent the height and width of the wood blocks, respectively.
Python3
class Solution:
def sellingWood(self, m: int, n: int, prices: List[List[int]]) -> int:
@cache
def dfs(h: int, w: int) -> int:
ans = d[h].get(w, 0)
for i in range(1, h // 2 + 1):
ans = max(ans, dfs(i, w) + dfs(h - i, w))
for i in range(1, w // 2 + 1):
ans = max(ans, dfs(h, i) + dfs(h, w - i))
return ans
d = defaultdict(dict)
for h, w, p in prices:
d[h][w] = p
return dfs(m, n)
Java
class Solution {
private int[][] d;
private Long[][] f;
public long sellingWood(int m, int n, int[][] prices) {
d = new int[m + 1][n + 1];
f = new Long[m + 1][n + 1];
for (var p : prices) {
d[p[0]][p[1]] = p[2];
}
return dfs(m, n);
}
private long dfs(int h, int w) {
if (f[h][w] != null) {
return f[h][w];
}
long ans = d[h][w];
for (int i = 1; i < h / 2 + 1; ++i) {
ans = Math.max(ans, dfs(i, w) + dfs(h - i, w));
}
for (int i = 1; i < w / 2 + 1; ++i) {
ans = Math.max(ans, dfs(h, i) + dfs(h, w - i));
}
return f[h][w] = ans;
}
}
C++
class Solution {
public:
long long sellingWood(int m, int n, vector<vector<int>>& prices) {
using ll = long long;
ll f[m + 1][n + 1];
int d[m + 1][n + 1];
memset(f, -1, sizeof(f));
memset(d, 0, sizeof(d));
for (auto& p : prices) {
d[p[0]][p[1]] = p[2];
}
function<ll(int, int)> dfs = [&](int h, int w) -> ll {
if (f[h][w] != -1) {
return f[h][w];
}
ll ans = d[h][w];
for (int i = 1; i < h / 2 + 1; ++i) {
ans = max(ans, dfs(i, w) + dfs(h - i, w));
}
for (int i = 1; i < w / 2 + 1; ++i) {
ans = max(ans, dfs(h, i) + dfs(h, w - i));
}
return f[h][w] = ans;
};
return dfs(m, n);
}
};
Go
func sellingWood(m int, n int, prices [][]int) int64 {
f := make([][]int64, m+1)
d := make([][]int, m+1)
for i := range f {
f[i] = make([]int64, n+1)
for j := range f[i] {
f[i][j] = -1
}
d[i] = make([]int, n+1)
}
for _, p := range prices {
d[p[0]][p[1]] = p[2]
}
var dfs func(int, int) int64
dfs = func(h, w int) int64 {
if f[h][w] != -1 {
return f[h][w]
}
ans := int64(d[h][w])
for i := 1; i < h/2+1; i++ {
ans = max(ans, dfs(i, w)+dfs(h-i, w))
}
for i := 1; i < w/2+1; i++ {
ans = max(ans, dfs(h, i)+dfs(h, w-i))
}
f[h][w] = ans
return ans
}
return dfs(m, n)
}
TypeScript
function sellingWood(m: number, n: number, prices: number[][]): number {
const f: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(-1));
const d: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
for (const [h, w, p] of prices) {
d[h][w] = p;
}
const dfs = (h: number, w: number): number => {
if (f[h][w] !== -1) {
return f[h][w];
}
let ans = d[h][w];
for (let i = 1; i <= Math.floor(h / 2); i++) {
ans = Math.max(ans, dfs(i, w) + dfs(h - i, w));
}
for (let i = 1; i <= Math.floor(w / 2); i++) {
ans = Math.max(ans, dfs(h, i) + dfs(h, w - i));
}
return (f[h][w] = ans);
};
return dfs(m, n);
}
Solution 2: Dynamic Programming
We can transform the memoization search in Solution 1 into dynamic programming.
Similar to Solution 1, we define a 2D array $d$, where $d[i][j]$ represents the price of a wood block with height $i$ and width $j$. Initially, we iterate through the price array $prices$ and store the price $p$ of each wood block $(h, w, p)$ in $d[h][w]$, while the rest of the prices are set to $0$.
Then, we define another 2D array $f$, where $f[i][j]$ represents the maximum amount of money obtained by cutting a wood block with height $i$ and width $j$. The answer will be $f[m][n]$.
Considering how $f[i][j]$ transitions, initially $f[i][j] = d[i][j]$. We enumerate the cutting positions, calculate the maximum amount of money obtained by cutting the wood block into two pieces, and take the maximum value.
The time complexity is $O(m \times n \times (m + n) + p)$, and the space complexity is $O(m \times n)$. Here, $p$ represents the length of the price array, while $m$ and $n$ represent the height and width of the wood blocks, respectively.
Similar problems:
Python3
class Solution:
def sellingWood(self, m: int, n: int, prices: List[List[int]]) -> int:
d = defaultdict(dict)
for h, w, p in prices:
d[h][w] = p
f = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
f[i][j] = d[i].get(j, 0)
for k in range(1, i):
f[i][j] = max(f[i][j], f[k][j] + f[i - k][j])
for k in range(1, j):
f[i][j] = max(f[i][j], f[i][k] + f[i][j - k])
return f[m][n]
Java
class Solution {
public long sellingWood(int m, int n, int[][] prices) {
int[][] d = new int[m + 1][n + 1];
long[][] f = new long[m + 1][n + 1];
for (int[] p : prices) {
d[p[0]][p[1]] = p[2];
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
f[i][j] = d[i][j];
for (int k = 1; k < i; ++k) {
f[i][j] = Math.max(f[i][j], f[k][j] + f[i - k][j]);
}
for (int k = 1; k < j; ++k) {
f[i][j] = Math.max(f[i][j], f[i][k] + f[i][j - k]);
}
}
}
return f[m][n];
}
}
C++
class Solution {
public:
long long sellingWood(int m, int n, vector<vector<int>>& prices) {
long long f[m + 1][n + 1];
int d[m + 1][n + 1];
memset(f, -1, sizeof(f));
memset(d, 0, sizeof(d));
for (auto& p : prices) {
d[p[0]][p[1]] = p[2];
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
f[i][j] = d[i][j];
for (int k = 1; k < i; ++k) {
f[i][j] = max(f[i][j], f[k][j] + f[i - k][j]);
}
for (int k = 1; k < j; ++k) {
f[i][j] = max(f[i][j], f[i][k] + f[i][j - k]);
}
}
}
return f[m][n];
}
};
Go
func sellingWood(m int, n int, prices [][]int) int64 {
d := make([][]int, m+1)
f := make([][]int64, m+1)
for i := range d {
d[i] = make([]int, n+1)
f[i] = make([]int64, n+1)
}
for _, p := range prices {
d[p[0]][p[1]] = p[2]
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
f[i][j] = int64(d[i][j])
for k := 1; k < i; k++ {
f[i][j] = max(f[i][j], f[k][j]+f[i-k][j])
}
for k := 1; k < j; k++ {
f[i][j] = max(f[i][j], f[i][k]+f[i][j-k])
}
}
}
return f[m][n]
}
TypeScript
function sellingWood(m: number, n: number, prices: number[][]): number {
const f: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
const d: number[][] = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
for (const [h, w, p] of prices) {
d[h][w] = p;
}
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
f[i][j] = d[i][j];
for (let k = 1; k < i; k++) {
f[i][j] = Math.max(f[i][j], f[k][j] + f[i - k][j]);
}
for (let k = 1; k < j; k++) {
f[i][j] = Math.max(f[i][j], f[i][k] + f[i][j - k]);
}
}
}
return f[m][n];
}