Appleman and Easy Task-cf_b - appleman and easy task-优快云博客

本文链接：https://blog.youkuaiyun.com/mids_ummer/article/details/38926167

本博客提供了解决一个棋盘问题的方法，即判断每个单元格周围是否有偶数个'O'字符。通过实现DFS算法，我们能够遍历棋盘并检查每个单元格的条件是否满足。

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Toastman came up with a very easy task. He gives it to Appleman, but Appleman doesn't know how to solve it. Can you help him?

Given a n × n checkerboard. Each cell of the board has either character 'x', or character 'o'. Is it true that each cell of the board has even number of adjacent cells with 'o'? Two cells of the board are adjacent if they share a side.

Input

The first line contains an integer n (1 ≤ n ≤ 100). Then n lines follow containing the description of the checkerboard. Each of them contains n characters (either 'x' or 'o') without spaces.

Output

Print "YES" or "NO" (without the quotes) depending on the answer to the problem.

Sample test(s)

input
3
xxo
xox
oxx

output
YES

input
4
xxxo
xoxo
oxox
xxxx

output
NO

我题目没注意到是不管x,还是o,都找寻周围是否有偶数个o

#include<iostream>
using namespace std;
char a[200][200];
int vis[200][200];
int n;
int dfs(int i, int j)
{
	int num = 0;
	if (a[i - 1][j] == 'o'&&i>0 && i <= n&&j <= n&&j>0)
		num++;
	if (a[i][j - 1] == 'o'&&i>0 && i <= n&&j <= n&&j>0)
		num++;
	if (a[i][j + 1] == 'o'&&i>0 && i <= n&&j <= n&&j>0)
		num++;
	if (a[i + 1][j] == 'o'&&i>0 && i <= n&&j <= n&&j>0)
		num++;
	return num;
}
int main()
{
	//int n;
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			cin >> a[i][j];
		}
	}
	int x = 0, flag = 0;
	for (int i = 1; i <= n; i++)
	{

		for (int j = 1; j <= n; j++)
		{
			
				x = dfs(i, j);
				if (x % 2 != 0)
				{
					flag = 1;
					break;
				}
			
		}
		if (flag)
			break;
	}
	
	if (flag)
		cout << "NO" << endl;
	else
		cout << "YES" << endl;
}