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题目描述


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?

输入

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

输出

One line for each test case.

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

样例输入

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

样例输出

Yes
No
No

提示

zoj2850

代码如下

#include <iostream>
#include <cstdio>
#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,mark;
bool Bound(int x,int y)
{
    return (x>=0&&y>=0&&x<n&&y<m);
}

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=1;
                        return ;
                    }
                }
            }
        }
}

int main()
{
    while(cin>>n>>m)
    {
        if(n==0&&m==0)
            break;
        int i,j;
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                cin>>map[i][j];
        mark=0;
        sum=0;
        Decide();
        if(mark||sum==m*n)
            cout<<"No"<<endl;
        else
            cout<<"Yes"<<endl;
    }
    return 0;
}

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