JAVA 数组代替QUEUE版

因为开辟数组，new的操作花费的代价系数接近100 所以这并不是一种好办法，使用java.util.LinkedList;来实现Queue接口更好，内部会有优化哦。但是我比较懒啦，。。值得注意的是，在java中不能像 c/c++ 一样使用&，在JAVA中int等基本数据类型是值传递，故此将bund和land放到方法内部比较方便，当然将他们放到全局静态区也行

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {   // flood fill

    //n = 10^3 共 10^6 个输入
    static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    static final int N = 1010;
    static int n, res;
    static char g[][] = new char[N][N];
    static boolean st[][] = new boolean[N][N];
    static int q[][] = new int[N * N][2];

    static int dx[] = {-1, 0, 1, 0};
    static int dy[] = {0, 1, 0, -1};

    static boolean bfs(int[] start) {
        int front = 0, rear = 0;
        q[0] = start;  //起点入队
        int land = 0, bund = 0;
        while (front <= rear) {
            int[] t = q[front ++ ];
            st[t[0]][t[1]] = true;  //标记访问过
            boolean is_bund = false;
            for (int i = 0; i < 4; i++) {
                int x = t[0] + dx[i], y = t[1] + dy[i];
                if(x < 0 || x >= n || y < 0 || y >= n) continue;
                if(st[x][y]) continue;
                if(g[x][y] == '.') {
                    is_bund = true;
                    continue;
                }
                land ++ ;
                q[++ rear] = new int[] {x, y};  //拓展队列
            }
            if (is_bund) bund ++ ;
        }

        return land == bund;
    }
    public static void main(String[] args) throws NumberFormatException, IOException {
        n = Integer.parseInt(in.readLine());
        for (int i = 0; i < n; i++) {
            g[i] = in.readLine().toCharArray();
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (!st[i][j] && g[i][j] == '#') {
                    if(bfs(new int[] {i, j})) res ++ ;
                }
            }
        }
        System.out.println(res);
    }

}

欢迎点评hhh