// 1970. Last Day Where You Can Still Cross #include "leetcode.h" /* * there is a 1-based binary matrix where 0 represents land and 1 represents * water. you are given integers 'row' and 'col' representing the number of rows * and columns in the matrix, respectively. initially on day 0, the entire * matrix is land. however each day a new cell becomes flooded with water. you * are given a 1-based 2d array 'cells', where 'cells[i] = [ri, ci]' represnts * that on the i'th day, the cell on the r[i'th] row and c[i'th] column (1-based * coordinates) will be covered with water (ie. changed to 1). you want to find * the last day that it is possible to walk from the top to the bottom by only * walking on land cells. you can start from any cell in the top row and end in * any cell in the bottom row. you can only travel in the four cardinal * directions (left, right, up, down) return the last day where it is possible * to walk from the top to the bottom by only walking on land cells */ class Solution { public: int latestDayToCross(int row, int col, vvd & cells) { vvd grid(row, vector(col, 1)); p = vector(row * col + 3, -1); int start_g = row * col + 1, end_g = row * col + 2; for (int i = 0; i < col; i++) { uni(i, start_g); uni((row - 1) * col + i, end_g); } for (int i = row * col - 1; i >= 0; i--) { int r = cells[i][0] - 1, c = cells[i][1] - 1; grid[r][c] = 0; int key = r * col + c; for (int k = 0; k < 4; ++k) { int n_r = r + dir[k][0], n_c = c + dir[k][1]; if (n_r >= 0 && n_r < row && n_c >= 0 && n_c < col && !grid[n_r][n_c]) { int n_key = n_r * col + n_c; uni(key, n_key); } } } return row * col - 1; } private: vector p; const int dir[4][2] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}}; int find(int x) { return p[x] < 0 ? x : p[x] = find(p[x]); } void uni(int x, int y) { x = find(x), y = find(y); if (x == y) return; p[y] = x; } }; int main() { Solution obj; vvd c1 = {{1, 1}, {2, 1}, {1, 2}, {2, 2}}, c2 = {{1, 1}, {1, 2}, {2, 1}, {2, 2}}; printf("%d\n", obj.latestDayToCross(2, 2, c1)); // expect: 2 printf("%d\n", obj.latestDayToCross(2, 2, c2)); // expect: 1 }