1162. As Far from Land as Possible

问题：

给定n*n二维数组，

0代表海水，1代表陆地，

求离陆地最远的海水cell，距离陆地多远。

Example 1:
Input: grid = [[1,0,1],[0,0,0],[1,0,1]]
Output: 2
Explanation: The cell (1, 1) is as far as possible from all the land with distance 2.

Example 2:
Input: grid = [[1,0,0],[0,0,0],[0,0,0]]
Output: 4
Explanation: The cell (2, 2) is as far as possible from all the land with distance 4.
 
Constraints:
n == grid.length
n == grid[i].length
1 <= n <= 100
grid[i][j] is 0 or 1

　　

example 1:

 

 

example 2:

 

 

解法：BFS

状态：cell坐标+当前坐标访问过？grid==1?

扩展层数即为所求最远距离。

base：首先将陆地，grid=1的cell加入queue。

对于每个状态node：遍历上下左右四个节点：

在grid范围内&&grid==0（未访问过&&是海水），则入队。且标记访问grid=1。

 

代码参考：

class Solution {
public:
    int dir[5] = {1,0,-1,0,1};
    int maxDistance(vector<vector<int>>& grid) {
        queue<pair<int,int>> q;
        int n = grid.size();
        for(int i=0; i<n; i++) {
            for(int j=0; j<n; j++) {
                if(grid[i][j]==1) {
                    q.push({i,j});
                }
            }
        }
        int step = 0;
        if(q.size()==n*n) return -1;
        while(!q.empty()) {
            int sz = q.size();
            for(int k=0; k<sz; k++) {
                auto [i,j] = q.front();
                q.pop();
                for(int d=1; d<5; d++) {
                    int x = i+dir[d-1];
                    int y = j+dir[d];
                    if(x<0 || y<0 || x>=n || y>=n || grid[x][y] != 0) continue;
                    q.push({x,y});
                    grid[x][y]=1;
                }
            }
            step++;
        }
        return step-1;
    }
};