ZOJ 2850:Beautiful Meadow

最新推荐文章于 2019-08-10 13:10:38 发布

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本篇介绍了一个关于Tom's Meadow的问题，通过给定的条件判断草地是否美丽。主要内容涉及输入输出处理及简单的逻辑判断算法。

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

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

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 Mintegers 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

题解：0代表被割，1代表有草，adjacent邻，mowed割。漂亮草地的条件：1不是所有方块都是草；2被割的不相邻。告诉自己以后输出跟操作还是分开写比较好......

代码：

#include <iostream>
using namespace std;
int a[15][15];
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        if(n==0&&m==0)
            break;
        int cnt=0;
        int flag=0;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                cin>>a[i][j];
            }
        }
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {

                if(a[i][j]==1)
                    cnt++;
                if(cnt==n*m)
                    flag=1;
                if(a[i][j]==0)
                {
                    if((j+1)<m&&a[i][j+1]==0)
                        flag=1;
                    else if(i-1>=0&&a[i-1][j]==0)
                        flag=1;
                    else if(j-1>=0&&a[i][j-1]==0)
                        flag=1;
                    else if(i+1<n&&a[i+1][j]==0)
                        flag=1;
                }

            }
        }
        //cout<<"cnt"<<cnt<<endl;
        //cout<<"flag"<<flag<<endl;
        if(flag==0)
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}