uva 439 Knight Moves(骑士的移动)BFS

最新推荐文章于 2021-05-13 21:02:10 发布

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最新推荐文章于 2021-05-13 21:02:10 发布

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本文链接：https://blog.youkuaiyun.com/jiaomenglei/article/details/61632099

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本文介绍了一种使用广度优先搜索(BFS)算法解决骑士在棋盘上从起点到终点最少步数的问题。通过定义骑士移动的八个方向，并利用队列进行节点扩展，最终输出了从指定起点到达终点所需的最短移动步数。

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BFS

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

public class Main {
	
	static int[] dr = {-2,-2,-1,1,2,2,1,-1};
	static int[] dc = {-1,1,2,2,1,-1,-2,-2};
	static boolean[][] vis = new boolean[9][9];
	static int ans;
	static int start_x,start_y,end_x,end_y;
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		while(scan.hasNext()){
			for(int i=1;i<=8;i++){
				Arrays.fill(vis[i], false);
			}
			String start = scan.next();
			String end = scan.next();
			start_x = start.charAt(0)-'a'+1;
			start_y = 9-(start.charAt(1)-'0');
			end_x = end.charAt(0)-'a'+1;
			end_y = 9-(end.charAt(1)-'0'); 
			Queue<Node> q = new LinkedList<>();
			Node u = new Node(start_x,start_y,0);
			q.add(u);
			while(!q.isEmpty()){
				u = q.poll();
				if(u.r==end_x&&u.c==end_y){
					System.out.printf("To get from %s to %s takes %d knight moves.\n",start,end,u.d);
					break;
				}
				for(int i=0;i<8;i++){
					int r = u.r+dr[i];
					int c = u.c+dc[i];
					if(inside(r,c)){
						q.add(new Node(r,c,u.d+1));
					}
				}
			}
		}
	}
	
	public static boolean inside(int r,int c){
		return r>=1&&r<=8&&c>=1&&c<=8;
	}

	static class Node{
		int r,c,d;
		public Node(int r,int c,int d){
			this.r = r;
			this.c = c;
			this.d = d;
		}
	}
}