《ACM程序设计》书中题目W

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本文介绍了一个判定Tom's Meadow是否美丽的算法。该算法通过检查草地中被修剪的部分是否相邻及是否存在未修剪区域来判断。输入为草地的布局矩阵，输出则为是否符合美丽标准。

Description

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 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

统计0的个数以及搜索是否存在纵向和横向两个方向之一存在连续0的情况，用二位map进行搜索。

代码如下：

#include<bits/stdc++.h>
using namespace std;
const int MAX = 12;
int main()
{
    int map[MAX][MAX];
    int i,j,n,m,flag1,flag2;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        if(n==0||m==0)
        {
            break;
        }
        for(i=0;i<MAX;++i)
        {
            for(j=0;j<MAX;++j)
            {
                map[i][j]=1;
            }
        }
        flag1=0;
        flag2=1;
        for(i=1;i<=n;++i)
        {
            for(j=1;j<=m;++j)
            {
                scanf("%d",&map[i][j]);
                if(map[i][j] == 0)
                {
                    flag1=1;
                }
            }
        }
        for(i=1;i<=n&&flag2;++i)
        {
            for(j=1;j<=m&&flag2;++j)
            {
                if((map[i-1][j]==0||map[i+1][j]==0||map[i][j+1]==0||map[i][j-1]==0)&&map[i][j]==0)
                {
                    flag2=0;

                }
            }
        }
        if(flag1&&flag2)
        {
            printf("Yes\n");
        } 
        else
        {
            printf("No\n");
        }     
    }
    return 0;
}