AcWing 1101. 献给阿尔吉侬的花束 宽搜 边界处理 java

原题地址


输入样例

3
3 4
.S..
###.
..E.
3 4
.S..
.E..
....
3 4
.S..
####
..E.

输出样例

5
1
oop!

⭐ 从起点开始，暴力BFS，用队列实现，走过的地方改成 墙 避免走老路

🤬 细节输出：使用 buffer 流输出记得手动打 换行符

package algorithm.lanQiao.宽搜;

import java.io.*;
import java.net.Inet4Address;
import java.util.*;

public class 献给阿尔吉侬的花束
{
	static BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
	static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));

	static int N = 210;
	static char[][] arr = new char[N][N];// 存地图
	static int[][] dist = new int[N][N];// 存距离
//  偏移量数组
	static int[] dx = { 0, 1, 0, -1 };
	static int[] dy = { 1, 0, -1, 0 };

	static class Pair
	{
		int x;// 存横坐标
		int y;// 存纵坐标

		public Pair(int x, int y)
		{
			super();
			this.x = x;
			this.y = y;
		}
	}

	public static void main(String[] args) throws IOException
	{
		int T = Integer.parseInt(in.readLine());
		while (T-- > 0)
		{
			String[] ss = in.readLine().split(" ");
			int r = Integer.parseInt(ss[0]);
			int c = Integer.parseInt(ss[1]);

//			初始化地图（整个地图都是墙）（宽搜的时候就不用特殊处理边界的问题了）
			for (int i = 0; i < N; i++)
			{
				Arrays.fill(arr[i], '#');
			}
//			初始化距离
			for (int i = 0; i < dist.length; i++)
			{
				Arrays.fill(dist[i], 0);
			}

			Pair start = null, end = null;// 用 Pair 记录起终点
//			输入
			for (int i = 1; i <= r; i++)
			{
				String s = in.readLine();
				for (int j = 1; j <= c; j++)
				{
					arr[i][j] = s.charAt(j - 1);
//					顺便记录起点和终点
					if (arr[i][j] == 'S')
						start = new Pair(i, j);
					if (arr[i][j] == 'E')
						end = new Pair(i, j);
				}
			}

			int res = bfs(start, end);
			if (res == -1)
				out.write("oop!\n");
			else
			{
				out.write(res + "\n");
			}
		}
		out.flush();
	}

	private static int bfs(Pair start, Pair end)
	{
		LinkedList<Pair> q = new LinkedList<>();
		q.add(start);

		while (!q.isEmpty())
		{
			Pair t = q.poll();
			int x = t.x;
			int y = t.y;

			for (int i = 0; i < 4; i++)
			{
				int xx = x + dx[i];
				int yy = y + dy[i];
				if (arr[xx][yy] == '#')
					continue;

				dist[xx][yy] = dist[x][y] + 1;
				if (arr[xx][yy] == 'E')
					return dist[xx][yy];
				arr[xx][yy] = '#';// 遍历过的地方改为墙，实现间接去重
				q.add(new Pair(xx, yy));
			}
		}
		return -1;
	}
}