1162 地图分析

https://leetcode-cn.com/problems/as-far-from-land-as-possible/submissions/

动态规划 dp[i][j]表示前i,j的最小值

int dpSearch(vector<vector<int>>& grid)
    {
        const int N = grid.size();
        const int INF = 200;
        int ans = -1;
        vector<vector<int>> dp(N, vector(N, INF));
        for(int r = 0; r < N; r++)
            for(int c = 0; c < N; c++)
            {
                if(grid[r][c]) dp[r][c] = 0;
                else dp[r][c] = min(r > 0 ? dp[r-1][c] : INF, c > 0 ? dp[r][c-1] : INF) + 1;
            }
        for(int r = N - 1; r >= 0; r--)
            for(int c = N - 1; c >= 0; c--)
            {
                if(r + 1 < N) dp[r][c] = min(dp[r][c], dp[r+1][c] + 1);
                if(c + 1 < N) dp[r][c] = min(dp[r][c], dp[r][c+1] + 1);
            } 
        for(int r = 0; r < N; r++)
            for(int c = 0; c < N; c++)
                if(dp[r][c]) ans = max(ans, dp[r][c]);   
        return ans >= INF? -1 : ans;
    }

BFS

int maxDistance(vector<vector<int>>& grid) {
        int n = grid.size();
        int dxy[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        queue<pair<int, int>> q;
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    q.push(make_pair(i, j));
                }
            }
        }
        if (q.empty() || q.size() == n * n) return -1;

        pair<int, int> cur;
        while (!q.empty()) {
            cur = q.front();
            q.pop();
            for (int i = 0; i < 4; i++) {
                int raw = cur.first + dxy[i][0];
                int column = cur.second + dxy[i][1];
                if (raw >= 0 && raw < n && column >= 0 && column < n && grid[raw][column] == 0) {
                    grid[raw][column] = grid[cur.first][cur.second] + 1;
                    q.push(make_pair(raw, column));
                }
            }
        }
        return grid[cur.first][cur.second] - 1;
    }