浙江省赛problem 1002（BF）

Tom's Meadow 美丽判定

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本题涉及一个简单的搜索算法，用于判断一个由1和0组成的M*N矩阵是否符合美丽草坪的标准：即矩阵中存在未修剪的草地且修剪过的草地不相邻。

Beautiful Meadow

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 1 Accepted Submission(s) : 2

Problem 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'reN lines each consisting ofM 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

Source

Zhejiang Provincial Programming Contest 2007

模拟水简单暴搜

题目大意：一块修剪过的草坪（M*N矩阵），有草则值为1，没草则值为0，看它是否beautiful。遵循两个原则 1、矩阵上这些点的值不全为1 2、两个值为0的点不能相邻

题目分析：控制好边界，然后上下左右地搜就行。

#include<stdio.h>
int main()
{
    int a[12][12],x,y,i,j,sum,flag;
    while(scanf("%d%d",&x,&y)!=EOF&&x&&y)
    {
        flag=sum=0;
        for(i=0;i<12;i++)//初始化
        {
            for(j=0;j<12;j++)
            {
                a[i][j]=2;//所有点为2
            }
        }
        for(i=1;i<=x;i++)//i和j都从1开始，好控制边界
        {
            for(j=1;j<=y;j++)
            {
                scanf("%d",&a[i][j]);
            }
        }
        for(i=1;i<=x;i++)
        {
            for(j=1;j<=y;j++)
            {
                sum+=a[i][j];
                if(a[i][j]==0)
                {
                    if(a[i][j-1]&&a[i][j+1]&&a[i-1][j]&&a[i+1][j])continue;
                    flag=1;
                }
            }
        }
        if(sum==x*y||flag)
        printf("No\n");
        else printf("Yes\n");
    }
    return 0;
}

本题核心算法：暴搜………………