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-rw-r--r--num_cut_pizza.cpp58
1 files changed, 0 insertions, 58 deletions
diff --git a/num_cut_pizza.cpp b/num_cut_pizza.cpp
deleted file mode 100644
index 526e21d..0000000
--- a/num_cut_pizza.cpp
+++ /dev/null
@@ -1,58 +0,0 @@
-// 1444. Number of Ways of Cutting a Pizza
-#include "leetcode.h"
-
-/*
- * given rectangular pizza represented as a rows x cols matrix containing the
- * following characters 'A' an apple and '.' empty cell, and given integer 'k'.
- * you have to cut the pizza into 'k' pieces using 'k - 1' cuts.
- */
-
-class Solution {
- void add(long &a, long &b) { a = (a + b) % (1e9 + 7); }
-
-public:
- int ways(vector<string> &pizza, int k) {
- int m = pizza.size(), n = pizza[0].size();
- vvd cnt(m + 1, vector<int>(n + 1));
- for (int i = m - 1; i >= 0; --i) {
- int s = 0;
- for (int j = n - 1; j >= 0; --j) {
- s += pizza[i][j] == 'A';
- cnt[i][j] = cnt[i + 1][j] + s;
- }
- }
- vvd(vector<long>) dp(m + 1, vvd(long)(n + 1, vector<long>(k + 1)));
- for (int i = m - 1; i >= 0; --i) {
- for (int j = n - 1; j >= 0; --j) {
- dp[i][j][1] = cnt[i][j] > 0;
- for (int p = 2; p <= k; ++p) {
- for (int q = i + 1; q < m; ++q) {
- if (cnt[i][j] == cnt[q][j])
- continue;
- if (cnt[q][j] == 0)
- break;
- add(dp[i][j][p], dp[q][j][p - 1]);
- }
- for (int q = j + 1; q < n; ++q) {
- if (cnt[i][j] == cnt[i][q])
- continue;
- if (cnt[i][q] == 0)
- break;
- add(dp[i][j][p], dp[i][q][p - 1]);
- }
- }
- }
- }
- return dp[0][0][k];
- }
-};
-
-int main() {
- Solution obj;
- vector<string> p1 = {"A..", "AAA", "..."};
- vector<string> p2 = {"A..", "AA.", "..."};
- vector<string> p3 = {"A..", "A..", "..."};
- printf("%d\n", obj.ways(p1, 3)); // expect: 3
- printf("%d\n", obj.ways(p2, 3)); // expect: 1
- printf("%d\n", obj.ways(p3, 1)); // expect: 1
-}