class Solution {
public:
int ways(vector<string>& pizza, int k) {
const int M = pizza.size();
const int N = pizza[0].size();
// dp[m][n][k] := # of ways to cut pizza[m:M][n:N] w/ k cuts
dp.resize(M, vector<vector<int>>(N, vector<int>(k, -1)));
prefix.resize(M + 1, vector<int>(N + 1));
for (int i = 0; i < M; ++i)
for (int j = 0; j < N; ++j)
prefix[i + 1][j + 1] = (pizza[i][j] == 'A') + prefix[i][j + 1] +
prefix[i + 1][j] - prefix[i][j];
return ways(0, 0, k - 1, M, N);
}
private:
constexpr static int kMod = 1e9 + 7;
vector<vector<vector<int>>> dp;
vector<vector<int>> prefix;
// hasApple of pizza[row1..row2)[col1..col2)
bool hasApple(int row1, int row2, int col1, int col2) {
return (prefix[row2][col2] - prefix[row1][col2] -
prefix[row2][col1] + prefix[row1][col1]) > 0;
};
int ways(int m, int n, int k, const int M, const int N) {
if (k == 0)
return 1;
if (dp[m][n][k] >= 0)
return dp[m][n][k];
dp[m][n][k] = 0;
for (int i = m + 1; i < M; ++i) // cut horizontally
if (hasApple(m, i, n, N) && hasApple(i, M, n, N))
dp[m][n][k] = (dp[m][n][k] + ways(i, n, k - 1, M, N)) % kMod;
for (int j = n + 1; j < N; ++j) // cut vertically
if (hasApple(m, M, n, j) && hasApple(m, M, j, N))
dp[m][n][k] = (dp[m][n][k] + ways(m, j, k - 1, M, N)) % kMod;
return dp[m][n][k];
}
};
leetcode
challenge
Here is the link for the problem:
https://leetcode.com/problems/number-of-ways-of-cutting-a-pizza/
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