Codeforces Round #263 (Div. 2) proA

解答Appleman的简单任务

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本文解释了如何解决Appleman提出的关于检查棋盘上每个单元格与其周围单元格中'o'字符数量是否全为偶数的简单问题。通过遍历棋盘并计算每个单元格周围'o'的数量，确定最终答案。

题目：

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

题意分析：

判断每个位置相邻位置为o的数字，如果全为偶数就yes，否者no。简单模拟题。Orz

代码：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <iostream>


using namespace std;

char a[105][105];
int main()
{
    int n;
    char c;
    int flag;
    cin >> n;
    memset(a, 0, sizeof(a));
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
        {
            cin >> c;
            if (c == 'o')
                a[i][j] = 1;
        }
    flag = 1;

    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
        {
            int cnt = 0;
            if (a[i-1][j])
                ++cnt;
            if (a[i+1][j])
                ++cnt;
            if (a[i][j-1])
                ++cnt;
            if (a[i][j+1])
                ++cnt;
            if (cnt & 1)
            {
                flag = false;
                break;
            }
        }
    if (flag)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;

    return 0;
}