Beautiful Meadow
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 1 Accepted Submission(s) : 2
Tom's Meadow
- 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
#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………………