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Diffstat (limited to 'last_day_cross.cpp')
-rw-r--r-- | last_day_cross.cpp | 62 |
1 files changed, 0 insertions, 62 deletions
diff --git a/last_day_cross.cpp b/last_day_cross.cpp deleted file mode 100644 index 4bf2e6b..0000000 --- a/last_day_cross.cpp +++ /dev/null @@ -1,62 +0,0 @@ -// 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<int>(col, 1)); - p = vector<int>(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<int> 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 -} |