zoj 2850 Beautiful Meadow

ZOJ Problem Set - 2850

Beautiful Meadow

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Tom's Meadow

Tom has a meadow in his garden. He divides it into N * M squares. Initially all the squares were covered with grass. He mowed down the grass on some of the squares and thinks the meadow is beautiful if and only if

1. Not all squares are covered with grass.
2. No two mowed squares are adjacent.

Two squares are adjacent if they share an edge. Here comes the problem: Is Tom's meadow beautiful now?

Input

The input contains multiple test cases!

Each test case starts with a line containing two integers N, M (1 <= N, M <= 10) separated by a space. There follows the description of Tom's Meadow. There're N lines each consisting of M integers separated by a space. 0(zero) means the corresponding position of the meadow is mowed and 1(one) means the square is covered by grass.

A line with N = 0 and M = 0 signals the end of the input, which should not be processed

Output

One line for each test case.

Output "Yes" (without quotations) if the meadow is beautiful, otherwise "No"(without quotations).

Sample Input

2 2
1 0
0 1
2 2
1 1
0 0
2 3
1 1 1
1 1 1
0 0

Sample Output

Yes
No
No

Author: CAO, Peng

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Source: Zhejiang Provincial Programming Contest 2007
Submit    Status

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// 1859941 2009-05-08 14:40:12 Accepted  2850 C++ 0 184 Wpl
// 1859935 2009-05-08 14:29:21 Wrong Answer  2850 C++ 0 184 Wpl  
#include  < iostream >
#define  MAX 12
using   namespace  std;
int  map[MAX][MAX];
int  a[ 4 ][ 2 ] = {{ 0 , 1 },{ 0 , - 1 },{ 1 , 0 },{ - 1 , 0 }};
int  n,m,sum;
bool  mark;
void  Init()
{
     int  i,j;
     for (i = 0 ;i < n;i ++ )
         for (j = 0 ;j < m;j ++ )
            scanf( " %d " , & map[i][j]);
}
bool  Bound( int  x, int  y)
{
     if (x >= 0 && y >= 0 && x < n && y < m)
         return   true ;
     else
         return   false ;
}
void  Decide()
{
     int  i,j,ti,tj,k;
     for (i = 0 ;i < n;i ++ )
         for (j = 0 ;j < m;j ++ )
        {
             if (map[i][j] == 1 )
            {
                sum ++ ;
                 continue ;
            }
             for (k = 0 ;k < 4 ;k ++ )
            {
                ti = i + a[k][ 0 ];   // 此处小心,绝不能写成i或者k，这里错过两次了
                tj = j + a[k][ 1 ];
                 if (Bound(ti,tj))
                {
                     if (map[ti][tj] == 0 )
                    {
                        mark = true ;
                         return  ;
                    }
                }
            }
        }
}
int  main()
{
     while (scanf( " %d%d " , & n, & m) != EOF)
    {
         if (n == 0 && m == 0 )
             break ;
        Init();
        mark = false ;
        sum = 0 ;
        Decide();
         if (mark || sum == m * n)
            printf( " No\n " );
         else
            printf( " Yes\n " );
    }
     return   0 ;
}