JAVA实现第九届蓝桥杯JAVA C组省赛 第三题 字母阵列

本文介绍了一个寻找特定字母序列“LANQIAO”在大型字母矩阵中的挑战,详细解释了序列可能存在的各种方向,并提供了一个示例算法解决方案。通过遍历矩阵的每一个位置,算法检查所有八个方向上的匹配情况,最终统计并输出所有符合条件的序列数量。

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字母阵列

仔细寻找,会发现:在下面的8x8的方阵中,隐藏着字母序列:“LANQIAO”。

SLANQIAO
ZOEXCCGB
MOAYWKHI
BCCIPLJQ
SLANQIAO
RSFWFNYA
XIFZVWAL
COAIQNAL

我们约定: 序列可以水平,垂直,或者是斜向;
并且走向不限(实际上就是有一共8种方向)。
上图中一共有4个满足要求的串。

下面有一个更大的(100x100)的字母方阵。
你能算出其中隐藏了多少个“LANQIAO”吗?

FOAIQNALWIKEGNICJWAOSXDHTHZPOLGYELORAUHOHCZIERPTOOJUITQJCFNIYYQHSBEABBQZPNGYQTCLSKZFCYWDGOAIADKLSNGJ
GSOZTQKCCSDWGUWAUOZKNILGVNLMCLXQVBJENIHIVLRPVVXXFTHQUXUAVZZOFFJHYLMGTLANQIAOQQILCDCJERJASNCTLYGRMHGF
TSDFYTLVIBHKLJVVJUDMKGJZGNNSTPVLCKTOFMUEUFSVQIAUVHNVFTGBDDARPKYNNCUOYUAZXQJNOEEYKLFRMOEMHUKJTPETHLES
FKVINSLWEVGAGMKVFVIUBMYOIAFHLVNKNTYKTZWVXQWRWIGPENFXYDTKRVPKRTHMGHVYOCLDCKMEKRLGEKBYUCLOLYPAKPFSOREH
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XWMYRTAFGFOCNHLBGNGOXPJSTWLZUNNAGIRETGXFWAQSSJPFTQAXMTQWMZWYVEPQERKSWTSCHSQOOBGXAQTBCHOEGBDVKGWJIFTG
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XZHDPZQTOBECJVBZLACVXACZEDYOGVAVQRNWEOWGRAQYUEUESTEDQTYJUTEFOOITSHDDZHONJGBRCWNEQLZUTBNQIADKNFIOMWZR
EBFKCVNLURZSNPOLTISRPDTNUMCDGKTYRGIOVEPTUTSBAWQKWWEUWIWHAANUZUADGZEATZOQICWFUJTWNZDBKLQNELWVTBNDLNFH
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SBAEKIFZCKDOMIQKBDWVQGBYWPDIBOLQUGAQRXLJDAZMXVZXYSNWEWTNZKYREMBEUHOTFOCKEJSXCMUBCKXNGQXTQJRCRCLWJTOI
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请提交一个整数,不要填写任何多余的内容。
解题思路:依次每个点进行八个方向的枚举,看看是否匹配"LANQIAO",如果匹配,则用一个二维数组标记一下,顺便计数,也可以不用二维数组标记,我就是好奇,想看一下,把符合"LANQIAO"所有方向都打印出来。输入方面可以自己用Scanner扫进去。答案是:41.

import java.util.Scanner;
public class Main
{
	static int n=100,count=0;//count计数
	static char [][]map;//接收,搜索
	static int [][]vis;//用于标记
	static char []word = "LANQIAO".toCharArray();//用于匹配
	static int []dx={0,1,0,-1,-1,-1,1,1};//用于存储八个方向
	static int []dy={1,0,-1,0,-1,1,-1,1};
	public static void main(String []args)
	{
		Scanner s = new Scanner(System.in);
		n = s.nextInt();
		map = new char[n][n];
		vis = new int [n][n];
		for(int x=0; x<n; x++)
		{
			map[x] = s.next().toCharArray();
		}
		for(int i=0; i<n; i++)//从0,0坐标开始对每个点进行搜索
		{
			for(int j=0; j<n; j++)
			{
				if(map[i][j] =='L')//从首字母开始
				{
					dp(i,j);
				}
			}
		}
		//result
		for(int x=0; x<n; x++)
		{
			for(int y =0; y<n; y++)
			{
				if(vis[x][y]==1)
				{
					System.out.print(map[x][y]);
				}
				else
				{
					System.out.print("*");
				}
			}
			System.out.println();
		}
		System.out.println(count);
	}
	public static void dp(int i,int j)
	{
				
		for(int x=0; x<8; x++)//从i,j该坐标开始,枚举它的八个方向
		{
				boolean flag = false;//标记
				int xx =i;
				int yy = j;
				for(int a=1; a<=6; a++)//从word的第二个字母开始循环匹配
				{
					xx +=dx[x];
			        yy +=dy[x];
					if(xx<0||xx>=n||yy<0||yy>=n||word[a]!=map[xx][yy])//如果角标越界or没匹配上,直接标记,以及break;
					{
						flag = true;
						break;
					}
				}
				if(!flag)//如果flag未被标记过,表示该方向符合"LANQIAO"
				{
					count++;//计数
					xx=i;//回到原始坐标
					yy=j;//回到原始坐标
					for(int c = 0; c<=6; c++)//对符合"LANQIAO"的某个方向用vis进行标记
					{
						vis[xx][yy] =1;
						xx+=dx[x];
						yy+=dy[x];
					}
				}
			
		}
	}
}

在这里插入图片描述

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