Codeforces 263A. Appleman and Easy Task

A. Appleman and Easy Task

time limit per test 

1 second

memory limit per test 

256 megabytes

input 

standard input

output 

standard output

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

解题：暴力枚举即可。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <climits>
 #include <vector>
 #include <queue>
 #include <cstdlib>
 #include <string>
 #include <set>
 #include <stack>
 #define LL long long
 #define pii pair<int,int>
 #define INF 0x3f3f3f3f
 using namespace std;
 const int maxn = ;
 int n;
 char mp[maxn][maxn];
 const int dir[][] = {,-,,,-,,,};
 int main() {
     int ans;
     bool flag;
     while(~scanf("%d",&n)){
         getchar();
         memset(mp,'x',sizeof(mp));
         for(int i = ; i <= n; i++){
             for(int j = ; j <= n; j++)
                 mp[i][j] = getchar();
             getchar();
         }
         flag = true;
         for(int i = ; i <= n; i++){
             for(int j = ; j <= n; j++){
                 ans = ;
                 for(int k = ; k < ; k++){
                     int x = i+dir[k][];
                     int y = j+dir[k][];
                     if(mp[x][y] == 'o') ans++;
                 }
                 if(ans&) {flag = false;break;}
             }
         }
         flag?puts("YES"):puts("NO");
     }
     return ;
 }