寒假打卡day4——1113. 红与黑

寒假打卡day4——1113. 红与黑

acwing 1113. 红与黑

Flood Fill 洪水灌溉算法

BFS

模板:

 while 队列非空{
     取出队头;
     枚举队头的四个邻格{
         if 各格子是陆地且为被开发过
            标记为开发过
            插入队列
     }
        
 }

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

/**
* @author JohnnyLin
* @version Creation Time:2021年1月14日 下午10:01:17
* 
*/
class Node{
	int x, y;
	public Node(int x, int y) {//注意赋值是最好用this.
		this.x = x;
		this.y = y;
	}
}
public class Main {
	//顺时针方向
	static int dir[][] = {{-1,0},{0,1},{1,0},{0,-1}};
	static int N = 25;
    static int m;
    static int n;
    static char[][] map = new char[N][N];
	public static int bfs(int sx,int sy) {
		int cnt = 1; 
		Queue<Node> q = new LinkedList<Node>();
		//标记为已经访问过
		q.add(new Node(sx, sy));
		//标记为红砖
		map[sx][sy] = '#';
		
		while(!	q.isEmpty()) {
			//取出队头
			Node node = q.peek();
			q.poll();
			
			//枚举队头的四个邻格
			for(int i = 0; i < 4; i++) {
				int nx = node.x + dir[i][0];
				int ny = node.y + dir[i][1];
				
				//越界或者访问过或者不为黑砖
				if(nx < 0 || nx >= m || ny < 0 || ny >=n || map[nx][ny]!='.')
					continue;
				//应该在此处标记访问过
				map[nx][ny] = '#';
				q.offer(new Node(nx,ny));
				cnt ++;
			}
			
		}
		return cnt;
	}

	public static void main(String[] args) {
		Scanner reader = new Scanner(System.in);
		while(true) {
			n = reader.nextInt();		//列
			m = reader.nextInt();		//行	
			if( m ==0|| n == 0)
				break;
			int sx = 0, sy = 0;
			for(int i = 0; i < m; i++) {
				String s = reader.next();
				for(int j = 0; j < n; j++) {
					map [i][j] = s.charAt(j);
					if(map [i][j] == '@') {
						sx = i;
						sy = j;
					}
				}
			}	
			System.out.println(bfs(sx, sy) );

			
		}
	}


}



DFS

模板

dfs(x,y){
    将(x,y)标记为开发;
    枚举(x,y)的4个邻格;
    if 邻格可走
        dfs(邻格);
}

import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

/**
* @author JohnnyLin
* @version Creation Time:2021年1月14日 下午10:01:17
* 
*/

public class Main {
	//顺时针方向
	static int dir[][] = {{-1,0},{0,1},{1,0},{0,-1}};
	static int N = 25;
    static int m;
    static int n;
    static char[][] map = new char[N][N];
    static boolean[][] vis = new boolean[N][N];//记录该点是否遍历过
	public static int dfs(int sx,int sy) {
		int res = 1;
		map[sx][sy] = '#';

		//枚举(sx,sy)的四个邻格
		for(int i = 0; i < 4; i++) {
			//如果邻格可走
			int nx = sx + dir[i][0], ny = sy +dir[i][1];
			if(nx >= 0 && nx < m && ny >= 0 && ny < n && map[nx][ny] =='.') {
				res += dfs(nx, ny);
			}
			
		}
		return res;
	}

	public static void main(String[] args) {
		Scanner reader = new Scanner(System.in);
		while(true) {
			n = reader.nextInt();		//列
			m = reader.nextInt();		//行	
			if( m ==0|| n == 0)
				break;
			int sx = 0, sy = 0;
			for(int i = 0; i < m; i++) {
				String s = reader.next();
				for(int j = 0; j < n; j++) {
					map [i][j] = s.charAt(j);
					if(map [i][j] == '@') {
						sx = i;
						sy = j;
					}
				}
			}	
			System.out.println(dfs(sx, sy) );

			
		}
	}


}


上一篇:CodeCraft-20 (Div. 2) D. Nash Matrix 构造 + dfs


下一篇:牛客算法周周练7-题解