ZOJ 2850 Beautiful Meadow

Beautiful Meadow

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Time Limit: 2 Seconds      Memory Limit: 65536 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

扫描一遍，满足下面两个条件：

1：不能所有的都是1

2：不能有0相邻

#include <cstdio>
using namespace std;

int main()
{
    int n, m;
    int a[11][11];
    while (scanf("%d%d", &n, &m), n){
        for (int i = 0; i < n; i++){
            for (int j = 0; j < m; j++){
                scanf("%d", &a[i][j]);
            }
        }
        int flag1 = 1, flag2 = 0;
        for (int i = 0; i < n; i++){
            for (int j = 0; j < m; j++){
                if (a[i][j] == 0){
                    flag1 = 0;
                    if (i + 1 < n && a[i + 1][j] == 0)
                        flag2 = 1;
                    if (j + 1 < m && a[i][j + 1] == 0)
                        flag2 = 1;
                }
            }
        }
        if (flag1 || flag2)
            printf("No\n");
        else
            printf("Yes\n");
    }
    return 0;
}