Codeforces Round #263 (Div. 2) A.Appleman and Easy Task(坑)-优快云博客

本文深入解析并提供了解决复杂矩阵问题的方法，即检查每个位置的四面相邻元素中'o'的数量是否为偶数。通过实例演示，帮助读者理解并解决类似Appleman遇到的EasyTask。

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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
 
 
 题解：
 个人感觉昨晚上的CF的A题坑到要死了，正确题意是若所给矩阵中的每一个位置四面相邻的共有偶数个'o'的话，就是YES，否则是NO，但真心无力吐槽题中的一句话Two cells of the board are adjacent if they share a side.什么是共用一条边。。。自然而然想到了同行或同列，但是WA了两个小时，今天看题解才知道，这句话就是个摆设，直接考虑边界判断str[i-1][j] str[i+1][j] str[i][j-1] str[i][j+1] 就好了。。。Orz。。。。。
 
 
 题目链接：http://codeforces.com/problemset/problem/462/A
 
 
 代码：
 
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;

int main(){
    int n;
    char str[110][110];
    scanf("%d",&n);
    getchar();
    for(int i=0;i<n;i++){
        gets(str[i]);
    }
    int ans=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            int cnt=0;
            if(i!=0&&str[i-1][j]=='o') cnt++;
            if(i!=n-1&&str[i+1][j]=='o') cnt++;
            if(j!=0&&str[i][j-1]=='o') cnt++;
            if(j!=n-1&&str[i][j+1]=='o') cnt++;
            if(cnt%2==1) ans=1;
        }
    }
    if(ans) printf("NO\n");
    else printf("YES\n");
    return 0;
}