题目这里

过得了样例和自己写的例子，但一提交就RE(用java写的)

import java.awt.Point;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;
import java.util.LinkedList;
import java.util.Queue;

public class Main {

	static int N, area = 0, cir = Integer.MAX_VALUE;
	static int[] xx = { 0, 0, 1, -1 };
	static int[] yy = { 1, -1, 0, 0 };
	static char[][] arr;
	static boolean[][] vis;

	public static void main(String[] args) throws IOException {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		StreamTokenizer st = new StreamTokenizer(bf);
		st.nextToken();
		N = (int) st.nval;
		arr = new char[N][N];
		vis = new boolean[N][N];
		bf.readLine();
		for (int i = 0; i < N; i++) {
			String str = bf.readLine();
			arr[i] = str.toCharArray();
		}
		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				if (!vis[i][j] && arr[i][j] == '#') {
					bfs(i, j, 1, 0);
				}
			}
		}
		System.out.println(area + " " + cir);
	}

	public static void bfs(int x, int y, int nowArea, int nowCir) {
		Queue<Point> q = new LinkedList<Point>();
		q.add(new Point(x, y));
		vis[x][y] = true;
		while (!q.isEmpty()) {
			Point p = q.poll();
			for (int i = 0; i < 4; i++) {
				int row = p.x + xx[i];
				int col = p.y + yy[i];
				if (row >= 0 && row < N && col >= 0 && col < N) {// 前提是不能越界
					if (!vis[row][col] && arr[row][col] == '#') {
						q.add(new Point(row, col));
						vis[row][col] = true;
						nowArea++;
					} else if (!vis[row][col] && arr[row][col] == '.') {
						nowCir++;
					}
				} else {
					nowCir++;
				}
			}
		}
		if (nowArea > area) {// 先要找到最大的
			area = nowArea;
			cir = nowCir;
		} else if (nowArea == area) {// 如果碰到相等的
			if (nowCir < cir) {// 就取周长最小的
				cir = nowCir;
			}
		}
	}
}