ZOJ-2850-Beautiful Meadow【暴力】【4th浙江省赛】

ZOJ-2850-Beautiful Meadow

                    Time Limit: 2 Seconds      Memory Limit: 65536 KB

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

Not all squares are covered with grass.
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

题目链接：ZOJ-2850

题目大意：给出一n*m只含有0 or 1的图。是否满足两个条件：1.不含有相邻的0； 2.存在0

题目思路：直接暴力

以下是代码：

#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <string>
#include <cstring>
using namespace std;
int a[20][20] = {0};
int n,m;
bool isbeauty()
{
    int flag = 0;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m - 1; j++)
        {
            if (a[i][j] == 0) flag = 1;
            if (a[i][j] == 0 && a[i][j + 1] == 0) return 0;
        }
    }
    for (int j = 0; j < m; j++)
    {
        for (int i = 0; i < n - 1; i++)
        {
            if (a[i][j] == 0) flag = 1;
            if (a[i][j] == 0 && a[i + 1][j] == 0) return 0;
        }
    }
    if (flag) return 1;
    else return 0;
}
int main(){
    while(cin >> n >> m)
    {
        if (n == 0 && m == 0) break;
        memset(a,0,sizeof(a));
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                cin >> a[i][j];
            }
        }
        int ans = isbeauty();
        if (ans) cout << "Yes\n";
        else cout << "No\n";
    }
    return 0;
}