// 2658. Maximum Number of Fish in a Grid #include "leetcode.h" /* * you are given a 0-indexed 2d matrix 'grid' of size 'm x n' where '(r, c)' * represents a land cell if 'grid[r][c] = 0', or a water cell containing * 'grid[r][c]' fish, if 'grid[r][c] > 0'. afisher can start at any water cell * and can do the following operations any number of times: catch all the fish * at the current cell, or move to any adjacent water cell. return the maximum * number of fish the fisher can catch if he chooses his starting cell * optimally, or 0 if no water cells exist. */ void dfs(int **grid, int r, int c, int n, int m, int *cnt) { if (r < 0 || r >= n || c < 0 || c >= m || grid[r][c] == 0) return; int dir[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}}; (*cnt) += grid[r][c]; grid[r][c] = 0; for (int i = 0; i < 4; i++) { int nr = r + dir[i][0]; int nc = c + dir[i][1]; dfs(grid, nr, nc, n, m, cnt); } } int findMaxFish(int **grid, int gridSize, int *gridColSize) { int ans = 0, cnt = 0; for (int i = 0; i < gridSize; i++) for (int j = 0; j < gridColSize[i]; j++) if (grid[i][j]) { dfs(grid, i, j, gridSize, gridColSize[i], &cnt); ans = fmax(ans, cnt); cnt = 0; } return ans; } int main() { int g1i[4][4] = {{0, 2, 1, 0}, {4, 0, 0, 3}, {1, 0, 0, 4}, {0, 3, 2, 0}}; int g2i[4][4] = {{1, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 1}}; struct two_d_arr g1, g2; two_d_arr_init(&g1, ARRAY_SIZE(g1i), ARRAY_SIZE(g1i[0]), g1i); two_d_arr_init(&g2, ARRAY_SIZE(g2i), ARRAY_SIZE(g2i[0]), g2i); printf("%d\n", findMaxFish(g1.arr, g1.row_size, g1.col_size)); // expect: 7 printf("%d\n", findMaxFish(g2.arr, g2.row_size, g2.col_size)); // expect: 1 two_d_arr_free(&g1); two_d_arr_free(&g2); }