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:
![](https://spcdn.pages.dev/leetcode/problems/2312.Selling%20Pieces%20of%20Wood/images/ex1.png)
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:
![](https://spcdn.pages.dev/leetcode/problems/2312.Selling%20Pieces%20of%20Wood/images/ex2new.png)
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 <= 200
1 <= prices.length <= 2 * 104
prices[i].length == 3
1 <= hi <= m
1 <= wi <= n
1 <= pricei <= 106
- All the shapes of wood
(hi, wi)
are pairwise distinct.
Solutions#
Solution 1#
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| class Solution:
def sellingWood(self, m: int, n: int, prices: List[List[int]]) -> int:
@cache
def dfs(h, w):
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)
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| class Solution {
private long[][] memo;
private int[][] d;
public long sellingWood(int m, int n, int[][] prices) {
d = new int[m + 1][n + 1];
memo = new long[m + 1][n + 1];
for (long[] e : memo) {
Arrays.fill(e, -1);
}
for (int[] p : prices) {
d[p[0]][p[1]] = p[2];
}
return dfs(m, n);
}
private long dfs(int m, int n) {
if (memo[m][n] != -1) {
return memo[m][n];
}
long ans = d[m][n];
for (int i = 1; i < m / 2 + 1; ++i) {
ans = Math.max(ans, dfs(i, n) + dfs(m - i, n));
}
for (int i = 1; i < n / 2 + 1; ++i) {
ans = Math.max(ans, dfs(m, i) + dfs(m, n - i));
}
memo[m][n] = ans;
return ans;
}
}
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| using ll = long long;
class Solution {
public:
long long sellingWood(int m, int n, vector<vector<int>>& prices) {
vector<vector<ll>> memo(m + 1, vector<ll>(n + 1, -1));
vector<vector<int>> d(m + 1, vector<int>(n + 1));
for (auto& p : prices) d[p[0]][p[1]] = p[2];
return dfs(m, n, d, memo);
}
ll dfs(int m, int n, vector<vector<int>>& d, vector<vector<ll>>& memo) {
if (memo[m][n] != -1) return memo[m][n];
ll ans = d[m][n];
for (int i = 1; i < m / 2 + 1; ++i) ans = max(ans, dfs(i, n, d, memo) + dfs(m - i, n, d, memo));
for (int i = 1; i < n / 2 + 1; ++i) ans = max(ans, dfs(m, i, d, memo) + dfs(m, n - i, d, memo));
memo[m][n] = ans;
return ans;
}
};
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| func sellingWood(m int, n int, prices [][]int) int64 {
memo := make([][]int, m+1)
d := make([][]int, m+1)
for i := range memo {
memo[i] = make([]int, n+1)
d[i] = make([]int, n+1)
for j := range memo[i] {
memo[i][j] = -1
}
}
for _, p := range prices {
d[p[0]][p[1]] = p[2]
}
var dfs func(int, int) int
dfs = func(m, n int) int {
if memo[m][n] != -1 {
return memo[m][n]
}
ans := d[m][n]
for i := 1; i < m/2+1; i++ {
ans = max(ans, dfs(i, n)+dfs(m-i, n))
}
for i := 1; i < n/2+1; i++ {
ans = max(ans, dfs(m, i)+dfs(m, n-i))
}
memo[m][n] = ans
return ans
}
return int64(dfs(m, n))
}
|
Solution 2#
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| 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
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
dp[i][j] = d[i].get(j, 0)
for k in range(1, i):
dp[i][j] = max(dp[i][j], dp[k][j] + dp[i - k][j])
for k in range(1, j):
dp[i][j] = max(dp[i][j], dp[i][k] + dp[i][j - k])
return dp[-1][-1]
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| class Solution {
public long sellingWood(int m, int n, int[][] prices) {
int[][] d = new int[m + 1][n + 1];
long[][] dp = 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) {
dp[i][j] = d[i][j];
for (int k = 1; k < i; ++k) {
dp[i][j] = Math.max(dp[i][j], dp[k][j] + dp[i - k][j]);
}
for (int k = 1; k < j; ++k) {
dp[i][j] = Math.max(dp[i][j], dp[i][k] + dp[i][j - k]);
}
}
}
return dp[m][n];
}
}
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| class Solution {
public:
long long sellingWood(int m, int n, vector<vector<int>>& prices) {
vector<vector<int>> d(m + 1, vector<int>(n + 1));
vector<vector<long long>> dp(m + 1, vector<long long>(n + 1));
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) {
dp[i][j] = d[i][j];
for (int k = 1; k < i; ++k) dp[i][j] = max(dp[i][j], dp[k][j] + dp[i - k][j]);
for (int k = 1; k < j; ++k) dp[i][j] = max(dp[i][j], dp[i][k] + dp[i][j - k]);
}
}
return dp[m][n];
}
};
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| func sellingWood(m int, n int, prices [][]int) int64 {
d := make([][]int, m+1)
dp := make([][]int, m+1)
for i := range d {
d[i] = make([]int, n+1)
dp[i] = make([]int, 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++ {
dp[i][j] = d[i][j]
for k := 1; k < i; k++ {
dp[i][j] = max(dp[i][j], dp[k][j]+dp[i-k][j])
}
for k := 1; k < j; k++ {
dp[i][j] = max(dp[i][j], dp[i][k]+dp[i][j-k])
}
}
}
return int64(dp[m][n])
}
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