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778. 水位上升的泳池中游泳

题目描述

在一个 n x n 的整数矩阵 grid 中，每一个方格的值 grid[i][j] 表示位置 (i, j) 的平台高度。

当开始下雨时，在时间为 t 时，水池中的水位为 t 。你可以从一个平台游向四周相邻的任意一个平台，但是前提是此时水位必须同时淹没这两个平台。假定你可以瞬间移动无限距离，也就是默认在方格内部游动是不耗时的。当然，在你游泳的时候你必须待在坐标方格里面。

你从坐标方格的左上平台 (0，0) 出发。返回 你到达坐标方格的右下平台 (n-1, n-1) 所需的最少时间 。

 

示例 1:

输入: grid = [[0,2],[1,3]]
输出: 3
解释:
时间为0时，你位于坐标方格的位置为 (0, 0)。
此时你不能游向任意方向，因为四个相邻方向平台的高度都大于当前时间为 0 时的水位。
等时间到达 3 时，你才可以游向平台 (1, 1). 因为此时的水位是 3，坐标方格中的平台没有比水位 3 更高的，所以你可以游向坐标方格中的任意位置

示例 2:

输入: grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
输出: 16
解释: 最终的路线用加粗进行了标记。
我们必须等到时间为 16，此时才能保证平台 (0, 0) 和 (4, 4) 是连通的

 

提示:

  • n == grid.length
  • n == grid[i].length
  • 1 <= n <= 50
  • 0 <= grid[i][j] < n2
  • grid[i][j] 中每个值 均无重复

解法

方法一

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class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        n = len(grid)
        p = list(range(n * n))
        hi = [0] * (n * n)
        for i, row in enumerate(grid):
            for j, h in enumerate(row):
                hi[h] = i * n + j
        for t in range(n * n):
            i, j = hi[t] // n, hi[t] % n
            for a, b in [(0, -1), (0, 1), (1, 0), (-1, 0)]:
                x, y = i + a, j + b
                if 0 <= x < n and 0 <= y < n and grid[x][y] <= t:
                    p[find(x * n + y)] = find(hi[t])
                if find(0) == find(n * n - 1):
                    return t
        return -1

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class Solution {
    private int[] p;

    public int swimInWater(int[][] grid) {
        int n = grid.length;
        p = new int[n * n];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        int[] hi = new int[n * n];
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                hi[grid[i][j]] = i * n + j;
            }
        }
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int t = 0; t < n * n; ++t) {
            int i = hi[t] / n;
            int j = hi[t] % n;
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k];
                int y = j + dirs[k + 1];
                if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] <= t) {
                    p[find(x * n + y)] = find(i * n + j);
                }
                if (find(0) == find(n * n - 1)) {
                    return t;
                }
            }
        }
        return -1;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

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class Solution {
public:
    vector<int> p;

    int swimInWater(vector<vector<int>>& grid) {
        int n = grid.size();
        p.resize(n * n);
        for (int i = 0; i < p.size(); ++i) p[i] = i;
        vector<int> hi(n * n);
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < n; ++j)
                hi[grid[i][j]] = i * n + j;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        for (int t = 0; t < n * n; ++t) {
            int i = hi[t] / n, j = hi[t] % n;
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] <= t)
                    p[find(x * n + y)] = find(hi[t]);
                if (find(0) == find(n * n - 1)) return t;
            }
        }
        return -1;
    }

    int find(int x) {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }
};

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func swimInWater(grid [][]int) int {
    n := len(grid)
    p := make([]int, n*n)
    for i := range p {
        p[i] = i
    }
    hi := make([]int, n*n)
    for i, row := range grid {
        for j, h := range row {
            hi[h] = i*n + j
        }
    }
    var find func(x int) int
    find = func(x int) int {
        if p[x] != x {
            p[x] = find(p[x])
        }
        return p[x]
    }
    dirs := []int{-1, 0, 1, 0, -1}
    for t := 0; t < n*n; t++ {
        i, j := hi[t]/n, hi[t]%n
        for k := 0; k < 4; k++ {
            x, y := i+dirs[k], j+dirs[k+1]
            if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] <= t {
                p[find(x*n+y)] = find(hi[t])
            }
            if find(0) == find(n*n-1) {
                return t
            }
        }
    }
    return -1
}

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function swimInWater(grid: number[][]): number {
    const m = grid.length,
        n = grid[0].length;
    let visited = Array.from({ length: m }, () => new Array(n).fill(false));
    let ans = 0;
    let stack = [[0, 0, grid[0][0]]];
    const dir = [
        [0, 1],
        [0, -1],
        [1, 0],
        [-1, 0],
    ];

    while (stack.length) {
        let [i, j] = stack.shift();
        ans = Math.max(grid[i][j], ans);
        if (i == m - 1 && j == n - 1) break;
        for (let [dx, dy] of dir) {
            let x = i + dx,
                y = j + dy;
            if (x < m && x > -1 && y < n && y > -1 && !visited[x][y]) {
                visited[x][y] = true;
                stack.push([x, y, grid[x][y]]);
            }
        }
        stack.sort((a, b) => a[2] - b[2]);
    }
    return ans;
}

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const DIR: [(i32, i32); 4] = [(-1, 0), (1, 0), (0, -1), (0, 1)];

impl Solution {
    #[allow(dead_code)]
    pub fn swim_in_water(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let m = grid[0].len();
        let mut ret_time = 0;
        let mut disjoint_set: Vec<usize> = vec![0; n * m];

        // Initialize the disjoint set
        for i in 0..n * m {
            disjoint_set[i] = i;
        }

        loop {
            if Self::check_and_union(&grid, &mut disjoint_set, ret_time) {
                break;
            }
            // Otherwise, keep checking
            ret_time += 1;
        }

        ret_time
    }

    #[allow(dead_code)]
    fn check_and_union(grid: &Vec<Vec<i32>>, d_set: &mut Vec<usize>, cur_time: i32) -> bool {
        let n = grid.len();
        let m = grid[0].len();

        for i in 0..n {
            for j in 0..m {
                if grid[i][j] != cur_time {
                    continue;
                }
                // Otherwise, let's union the square with its neighbors
                for (dx, dy) in DIR {
                    let x = dx + (i as i32);
                    let y = dy + (j as i32);
                    if Self::check_bounds(x, y, n as i32, m as i32)
                        && grid[x as usize][y as usize] <= cur_time
                    {
                        Self::union(i * m + j, (x as usize) * m + (y as usize), d_set);
                    }
                }
            }
        }

        Self::find(0, d_set) == Self::find(n * m - 1, d_set)
    }

    #[allow(dead_code)]
    fn find(x: usize, d_set: &mut Vec<usize>) -> usize {
        if d_set[x] != x {
            d_set[x] = Self::find(d_set[x], d_set);
        }
        d_set[x]
    }

    #[allow(dead_code)]
    fn union(x: usize, y: usize, d_set: &mut Vec<usize>) {
        let p_x = Self::find(x, d_set);
        let p_y = Self::find(y, d_set);
        d_set[p_x] = p_y;
    }

    #[allow(dead_code)]
    fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
        i >= 0 && i < n && j >= 0 && j < m
    }
}

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