leetcode-1020.飞地的数量

package com;

import java.util.Scanner;

public class Main {
	@SuppressWarnings("resource")
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		Solution s = new Solution();
		int T;
		T = scan.nextInt();
		while (T-- != 0) {
			int m = scan.nextInt();
			int n = scan.nextInt();
			int[][] A = new int[m][n];
			for (int i = 0; i < m; ++i)
				for (int j = 0; j < n; j++) {
					A[i][j] = scan.nextInt();
				}
			int res = s.numEnclaves(A);
			System.out.println(res);
		}
	}
}

class Solution {
	public int[][] vis;

	public int numEnclaves(int[][] A) {
		int row = A.length, col = A[0].length;
		vis = new int[row][col];
		for (int i = 0; i < row; ++i) {
			for (int j = 0; j < col; ++j) {
				if (i * j == 0 || i == row - 1 || j == col - 1) {
					A[i][j] = 2;
					vis[i][j] = 1;
					dfs(i, j, A);
				}
			}
		}
		int res = 0;
		for (int i = 0; i < row; ++i) {
			for (int j = 0; j < col; ++j) {
				if (A[i][j] == 1)
					++res;
			}
		}
		return res;
	}

	private void dfs(int x, int y, int[][] A) {
		// TODO Auto-generated method stub
		if (x < 0 || y < 0 || x >= A.length || y >= A[0].length || A[x][y] != 1)
			return;
		A[x][y] = 2;
		dfs(x + 1, y, A);
		dfs(x - 1, y, A);
		dfs(x, y + 1, A);
		dfs(x, y - 1, A);
	}
}
内部的1要想走出去,一定是一串1的路径
那么我们逆向思维,从边界的1向内部dfs,将经过的1感染为2
最后剩下的1就是被0包围1

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转载自blog.csdn.net/qq_39370495/article/details/89302123