Red and Black HDU 1312

题目链接点击这里Red and Black

使用方法：递归的深度优先遍历和非递归的深度优先遍历两种实现方式

代码：

import java.io.*;
import java.util.Stack;
import java.util.Scanner;

public class Main {
	private static Scanner in = new Scanner(System.in);
	private static char [][]matrix;
	private static int sum;
	private static int w;
	private static int h;
	private static int []a = {1,-1,0,0};
	private static int []b = {0,0,-1,1};
	private static boolean isLegal(int x,int y){
		if(x<0 || x>=h || y<0 || y>=w) return false;
		return true;
	}
//  递归的dfs
//	private static void dfs(int x,int y){
//		if(matrix[x][y]=='.') {
//			sum++;
//			matrix[x][y] = '*';
//		}
//		for(int i=0;i<4;i++)
//		{
//			int dx = x + a[i];
//			int dy = y + b[i];
//			if(judge(dx,dy) && matrix[dx][dy]=='.'){
//				dfs(dx,dy);
//			}
//		}
//	}
	private static class Point{
		int x,y;
		public Point(int x, int y){
			this.x = x;
			this.y = y;
		}
}
    //非递归的dfs
	private static void dfs(int x , int y){
		Stack<Point> stack = new Stack<>();
		stack.add(new Point(x,y));
		while(!stack.isEmpty()){
			Point p = stack.pop();
			x = p.x;
			y = p.y;
			if(matrix[x][y]=='.'){
				sum++;
				matrix[x][y] = '*';
			}
			for(int i=0; i<4; i++)
			{
				int dx = x + a[i];
				int dy = y + b[i];
				if(isLegal(dx,dy)){
					if(matrix[dx][dy]=='.'){
						stack.push(new Point(dx,dy));
					}
				}
			}
		}

	}
	public static void main(String[] args) throws IOException{
		w = in.nextInt();
		h = in.nextInt();
		String str;
		int x=0,y=0;
		while(w!=0 || h!=0){
			matrix = new char[h][w];
			for(int i=0;i<h;i++)
			{
				str = in.next();

				for(int j=0;j<w;j++)
				{
					matrix[i][j] = str.charAt(j);
					if(matrix[i][j]=='@'){
						x = i;
						y = j;
					}
				}
			}
			sum = 0;
			dfs(x,y);
			System.out.println(sum+1);
			w = in.nextInt();
			h = in.nextInt();
		}

	}
}