leetcode 778. Swim in Rising Water-优快云博客

本文介绍了一个算法问题：在一个NxN的网格中，每个格子的值代表该位置的高度。从左上角出发，在不同时间水位上升的情况下，如何找到到达右下角所需的最短时间。文章通过优先队列实现了一种有效的解决方案。

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On an N x N grid, each square grid[i][j] represents the elevation at that point (i,j).

Now rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square (0, 0). What is the least time until you can reach the bottom right square (N-1, N-1)?

Example 1:

Input: [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.

You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.

Example 2:

Input: [[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]]
Output: 16
Explanation:
 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

The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.

Note:

2 <= N <= 50.
grid[i][j] is a permutation of [0, ..., N*N - 1].

先占个坑，等我有空写注释

class Solution {
public:
    int swimInWater(vector<vector<int>>& grid) 
    {
        int n = grid.size();
        priority_queue<pair<int,int>> q;//pair<-time,y*n+x>;
        vector<int> visited(n*n);
        vector<int> dirs{-1,0,1,0,-1};
        q.push({-grid[0][0],0});
        visited[0]=1;
        while(!q.empty())
        {
            pair<int,int> node = q.top();
            q.pop();
            int time = -node.first;
            int x = node.second % n;
            int y = node.second / n;
            for(int i = 0;i<4;i++)
            {
                int tx = x + dirs[i];
                int ty = y + dirs[i+1];
                if(tx < 0 || ty < 0 || tx>=n || ty >= n) continue;
                if(visited[ty*n+tx]) continue;
                if(ty == n-1 && tx == n-1) return max(time,grid[n-1][n-1]);
                visited[ty*n+tx] = 1;
                q.push({-max(time,grid[ty][tx]),ty*n+tx});
            }
        }
        return -1;
    }
};