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feat: add solutions to lc problem: No.1444
No.1444.Number of Ways of Cutting a Pizza
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‎README.md

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- [网格图中递增路径的数目](/solution/2300-2399/2328.Number%20of%20Increasing%20Paths%20in%20a%20Grid/README.md) - `DFS`、`记忆化搜索`
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- [翻转游戏 II](/solution/0200-0299/0294.Flip%20Game%20II/README.md) - `DFS`、`状态压缩`、`记忆化搜索`
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- [统计所有可行路径](/solution/1500-1599/1575.Count%20All%20Possible%20Routes/README.md) - `DFS`、`记忆化搜索`
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- [切披萨的方案数](/solution/1400-1499/1444.Number%20of%20Ways%20of%20Cutting%20a%20Pizza/README.md) - `DFS`、`记忆化搜索`
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<!-- DFS 待补充 -->
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‎README_EN.md

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@@ -99,6 +99,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
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- [Number of Increasing Paths in a Grid](/solution/2300-2399/2328.Number%20of%20Increasing%20Paths%20in%20a%20Grid/README.md) - `DFS`、`Memoization`
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- [Flip Game II](/solution/0200-0299/0294.Flip%20Game%20II/README_EN.md) - `DFS`、`Bitmask`、`Memoization`
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- [Count All Possible Routes](/solution/1500-1599/1575.Count%20All%20Possible%20Routes/README_EN.md) - `DFS`、`Memoization`
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- [Number of Ways of Cutting a Pizza](/solution/1400-1499/1444.Number%20of%20Ways%20of%20Cutting%20a%20Pizza/README_EN.md) - `DFS`、`Memoization`
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### 4. Dynamic Programming(DP)
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‎solution/1400-1499/1444.Number of Ways of Cutting a Pizza/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：二维前缀和 + 记忆化搜索**
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时间复杂度 $O(mnk*(m+n))$。
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相似题目：[2312. 卖木头块](/solution/2300-2399/2312.Selling%20Pieces%20of%20Wood/README.md)
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def ways(self, pizza: List[str], k: int) -> int:
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@cache
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def dfs(i, j, k):
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if k == 0:
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return int(s[-1][-1] - s[-1][j] - s[i][-1] + s[i][j] > 0)
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res = 0
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for x in range(i + 1, m):
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if s[x][-1] - s[x][j] - s[i][-1] + s[i][j]:
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res += dfs(x, j, k - 1)
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for y in range(j + 1, n):
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if s[-1][y] - s[-1][j] - s[i][y] + s[i][j]:
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res += dfs(i, y, k - 1)
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return res % mod
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mod = 10**9 + 7
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m, n = len(pizza), len(pizza[0])
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s = [[0] * (n + 1) for _ in range(m + 1)]
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for i, row in enumerate(pizza):
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for j, v in enumerate(row):
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s[i + 1][j + 1] = s[i + 1][j] + \
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s[i][j + 1] - s[i][j] + int(v == 'A')
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return dfs(0, 0, k - 1)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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private static final int MOD = (int) 1e9 + 7;
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private int[][][] f;
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private int[][] s;
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private int m;
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private int n;
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public int ways(String[] pizza, int k) {
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m = pizza.length;
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n = pizza[0].length();
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s = new int[m + 1][n + 1];
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f = new int[m][n][k];
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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s[i + 1][j + 1] = s[i + 1][j] + s[i][j + 1] - s[i][j] + (pizza[i].charAt(j) == 'A' ? 1 : 0);
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Arrays.fill(f[i][j], -1);
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}
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}
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return dfs(0, 0, k - 1);
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}
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private int dfs(int i, int j, int k) {
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if (f[i][j][k] != -1) {
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return f[i][j][k];
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}
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if (k == 0) {
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return s[m][n] - s[m][j] - s[i][n] + s[i][j] > 0 ? 1 : 0;
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}
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int res = 0;
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for (int x = i + 1; x < m; ++x) {
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if (s[x][n] - s[x][j] - s[i][n] + s[i][j] > 0) {
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res = (res + dfs(x, j, k - 1)) % MOD;
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}
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}
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for (int y = j + 1; y < n; ++y) {
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if (s[m][y] - s[m][j] - s[i][y] + s[i][j] > 0) {
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res = (res + dfs(i, y, k - 1)) % MOD;
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}
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}
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f[i][j][k] = res;
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return res;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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const int mod = 1e9 + 7;
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vector<vector<vector<int>>> f;
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vector<vector<int>> s;
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int m;
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int n;
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int ways(vector<string>& pizza, int k) {
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m = pizza.size();
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n = pizza[0].size();
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s.assign(m + 1, vector<int>(n + 1, 0));
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f.assign(m, vector<vector<int>>(n, vector<int>(k, -1)));
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for (int i = 0; i < m; ++i)
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for (int j = 0; j < n; ++j)
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s[i + 1][j + 1] = s[i + 1][j] + s[i][j + 1] - s[i][j] + (pizza[i][j] == 'A');
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return dfs(0, 0, k - 1);
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}
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int dfs(int i, int j, int k) {
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if (f[i][j][k] != -1) return f[i][j][k];
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if (k == 0) return s[m][n] - s[m][j] - s[i][n] + s[i][j] > 0;
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int res = 0;
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for (int x = i + 1; x < m; ++x)
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if (s[x][n] - s[x][j] - s[i][n] + s[i][j])
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res = (res + dfs(x, j, k - 1)) % mod;
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for (int y = j + 1; y < n; ++y)
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if (s[m][y] - s[m][j] - s[i][y] + s[i][j])
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res = (res + dfs(i, y, k - 1)) % mod;
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f[i][j][k] = res;
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return res;
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}
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};
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```
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### **Go**
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```go
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func ways(pizza []string, k int) int {
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mod := int(1e9) + 7
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m, n := len(pizza), len(pizza[0])
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f := make([][][]int, m)
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s := make([][]int, m+1)
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for i := range f {
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f[i] = make([][]int, n)
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for j := range f[i] {
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f[i][j] = make([]int, k)
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for h := range f[i][j] {
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f[i][j][h] = -1
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}
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}
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}
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for i := range s {
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s[i] = make([]int, n+1)
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}
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for i, p := range pizza {
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for j, v := range p {
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s[i+1][j+1] = s[i+1][j] + s[i][j+1] - s[i][j]
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if v == 'A' {
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s[i+1][j+1]++
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}
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}
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}
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var dfs func(int, int, int) int
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dfs = func(i, j, k int) int {
210+
if f[i][j][k] != -1 {
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return f[i][j][k]
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}
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if k == 0 {
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if s[m][n]-s[m][j]-s[i][n]+s[i][j] > 0 {
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return 1
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}
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return 0
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}
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res := 0
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for x := i + 1; x < m; x++ {
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if s[x][n]-s[x][j]-s[i][n]+s[i][j] > 0 {
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res = (res + dfs(x, j, k-1)) % mod
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}
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}
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for y := j + 1; y < n; y++ {
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if s[m][y]-s[m][j]-s[i][y]+s[i][j] > 0 {
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res = (res + dfs(i, y, k-1)) % mod
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}
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}
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f[i][j][k] = res
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return res
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}
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return dfs(0, 0, k-1)
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}
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```
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### **...**

‎solution/1400-1499/1444.Number of Ways of Cutting a Pizza/README_EN.md

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### **Python3**
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```python
68-
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class Solution:
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def ways(self, pizza: List[str], k: int) -> int:
70+
@cache
71+
def dfs(i, j, k):
72+
if k == 0:
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return int(s[-1][-1] - s[-1][j] - s[i][-1] + s[i][j] > 0)
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res = 0
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for x in range(i + 1, m):
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if s[x][-1] - s[x][j] - s[i][-1] + s[i][j]:
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res += dfs(x, j, k - 1)
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for y in range(j + 1, n):
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if s[-1][y] - s[-1][j] - s[i][y] + s[i][j]:
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res += dfs(i, y, k - 1)
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return res % mod
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mod = 10**9 + 7
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m, n = len(pizza), len(pizza[0])
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s = [[0] * (n + 1) for _ in range(m + 1)]
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for i, row in enumerate(pizza):
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for j, v in enumerate(row):
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s[i + 1][j + 1] = s[i + 1][j] + \
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s[i][j + 1] - s[i][j] + int(v == 'A')
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return dfs(0, 0, k - 1)
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```
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### **Java**
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```java
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class Solution {
97+
private static final int MOD = (int) 1e9 + 7;
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private int[][][] f;
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private int[][] s;
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private int m;
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private int n;
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public int ways(String[] pizza, int k) {
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m = pizza.length;
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n = pizza[0].length();
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s = new int[m + 1][n + 1];
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f = new int[m][n][k];
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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s[i + 1][j + 1] = s[i + 1][j] + s[i][j + 1] - s[i][j] + (pizza[i].charAt(j) == 'A' ? 1 : 0);
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Arrays.fill(f[i][j], -1);
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}
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}
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return dfs(0, 0, k - 1);
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}
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private int dfs(int i, int j, int k) {
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if (f[i][j][k] != -1) {
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return f[i][j][k];
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}
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if (k == 0) {
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return s[m][n] - s[m][j] - s[i][n] + s[i][j] > 0 ? 1 : 0;
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}
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int res = 0;
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for (int x = i + 1; x < m; ++x) {
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if (s[x][n] - s[x][j] - s[i][n] + s[i][j] > 0) {
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res = (res + dfs(x, j, k - 1)) % MOD;
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}
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}
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for (int y = j + 1; y < n; ++y) {
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if (s[m][y] - s[m][j] - s[i][y] + s[i][j] > 0) {
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res = (res + dfs(i, y, k - 1)) % MOD;
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}
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}
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f[i][j][k] = res;
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return res;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
145+
public:
146+
const int mod = 1e9 + 7;
147+
vector<vector<vector<int>>> f;
148+
vector<vector<int>> s;
149+
int m;
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int n;
151+
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int ways(vector<string>& pizza, int k) {
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m = pizza.size();
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n = pizza[0].size();
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s.assign(m + 1, vector<int>(n + 1, 0));
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f.assign(m, vector<vector<int>>(n, vector<int>(k, -1)));
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for (int i = 0; i < m; ++i)
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for (int j = 0; j < n; ++j)
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s[i + 1][j + 1] = s[i + 1][j] + s[i][j + 1] - s[i][j] + (pizza[i][j] == 'A');
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return dfs(0, 0, k - 1);
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}
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int dfs(int i, int j, int k) {
164+
if (f[i][j][k] != -1) return f[i][j][k];
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if (k == 0) return s[m][n] - s[m][j] - s[i][n] + s[i][j] > 0;
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int res = 0;
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for (int x = i + 1; x < m; ++x)
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if (s[x][n] - s[x][j] - s[i][n] + s[i][j])
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res = (res + dfs(x, j, k - 1)) % mod;
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for (int y = j + 1; y < n; ++y)
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if (s[m][y] - s[m][j] - s[i][y] + s[i][j])
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res = (res + dfs(i, y, k - 1)) % mod;
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f[i][j][k] = res;
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return res;
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}
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};
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```
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### **Go**
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```go
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func ways(pizza []string, k int) int {
183+
mod := int(1e9) + 7
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m, n := len(pizza), len(pizza[0])
185+
f := make([][][]int, m)
186+
s := make([][]int, m+1)
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for i := range f {
188+
f[i] = make([][]int, n)
189+
for j := range f[i] {
190+
f[i][j] = make([]int, k)
191+
for h := range f[i][j] {
192+
f[i][j][h] = -1
193+
}
194+
}
195+
}
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for i := range s {
197+
s[i] = make([]int, n+1)
198+
}
199+
for i, p := range pizza {
200+
for j, v := range p {
201+
s[i+1][j+1] = s[i+1][j] + s[i][j+1] - s[i][j]
202+
if v == 'A' {
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s[i+1][j+1]++
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}
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}
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}
207+
var dfs func(int, int, int) int
208+
dfs = func(i, j, k int) int {
209+
if f[i][j][k] != -1 {
210+
return f[i][j][k]
211+
}
212+
if k == 0 {
213+
if s[m][n]-s[m][j]-s[i][n]+s[i][j] > 0 {
214+
return 1
215+
}
216+
return 0
217+
}
218+
res := 0
219+
for x := i + 1; x < m; x++ {
220+
if s[x][n]-s[x][j]-s[i][n]+s[i][j] > 0 {
221+
res = (res + dfs(x, j, k-1)) % mod
222+
}
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}
224+
for y := j + 1; y < n; y++ {
225+
if s[m][y]-s[m][j]-s[i][y]+s[i][j] > 0 {
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res = (res + dfs(i, y, k-1)) % mod
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}
228+
}
229+
f[i][j][k] = res
230+
return res
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}
232+
return dfs(0, 0, k-1)
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}
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```
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### **...**

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