1970. Last Day Where You Can Still Cross

題目描述

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] represents that on the i^th day, the cell on the ri^th row and ci^th column (1-based coordinates) will be covered with water (i.e., 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 at any cell in the bottom row. You can only travel in the four cardinal directions (left, right, up, and down).

Return the last day where it is possible to walk from the top to the bottom by only walking on land cells.

範例

Example 1:

Input: row = 2, col = 2, cells = [[1,1],[2,1],[1,2],[2,2]]
Output: 2
Explanation: The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 2.

Example 2:

Input: row = 2, col = 2, cells = [[1,1],[1,2],[2,1],[2,2]]
Output: 1
Explanation: The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 1.

Example 3:

Input: row = 3, col = 3, cells = [[1,2],[2,1],[3,3],[2,2],[1,1],[1,3],[2,3],[3,2],[3,1]]
Output: 3
Explanation: The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 3.

Constraints:

• 2 <= row, col <= 2 * 10⁴
• 4 <= row * col <= 2 * 10⁴
• cells.length == row * col
• 1 <= ri <= row
• 1 <= ci <= col
• All the values of cells are unique.

解答

C++

class Solution {
public:
    const int LAND = 0;
    const int WATER = 1;
    const int VISITED = 2;
    vector<pair<int, int>> delta = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    int latestDayToCross(int row, int col, vector<vector<int>>& cells) {
        ios_base::sync_with_stdio(0); cin.tie(0);
        int num_day = row * col;
        return binarySearch(num_day, row, col, cells);
    }

    int binarySearch(int num_day, int m, int n, vector<vector<int>>& cells) {
        int left = 0, right = num_day;
        while (left < right) {
            int mid = left + (right - left) / 2;
            bool reachable = isReachable(mid, m, n, cells);
            // cout << "\tmid = " << mid << "\treachable = " << reachable << endl;
            if (reachable) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return right - 1;
    }

    bool isReachable(int nth_day, int m, int n, vector<vector<int>>& cells) {
        vector<vector<int>> grid(m, vector<int>(n, LAND));
        queue<pair<int, int>> frontier;
        for (int i = 0; i < nth_day; i ++) {
            grid[cells[i][0] - 1][cells[i][1] - 1] = WATER;
        }
        for (int c = 0; c < n; c ++) {
            if (grid[0][c] == LAND) {
                frontier.push({0, c});
                grid[0][c] = VISITED;
            }
        }
        while (not frontier.empty()) {
            const auto [r, c] = frontier.front();
            frontier.pop();
            // printf("\t(%d, %d)", r, c);
            for (const auto [dr, dc] : delta) {
                if (r + dr < 0 or r + dr >= m or \
                    c + dc < 0 or c + dc >= n or \
                    grid[r + dr][c + dc] != LAND) {
                        continue;
                }
                grid[r + dr][c + dc] = VISITED;
                if (r + dr == m - 1) {
                    return true;
                }
                frontier.push({r + dr, c + dc});
            }
        }
        return false;
    }
};

Binary Search + BFS

Jerry Wu30 June, 2023

Reference

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