做梦都能笑醒的题解—简易暴力法

#include<cstdio>
using namespace std;
int a[110][110];
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {
            scanf("%d",&a[i][j]);        
        }
    }
    a[0][0] = 100001;
    for(int i = 0; i <= m+1; i++)
    {
        a[0][i] = 100001;
        a[n+1][i] = 100001;
    }
    for(int i = 0; i <= n+1; i++)
    {
        a[i][0] = 100001;
        a[i][m+1] = 100001;
    }
    int cnt = 0;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= m; j++)
        {
            if(a[i][j] <= a[i-1][j] && a[i][j] <= a[i+1][j] && a[i][j] <= a[i][j+1] && a[i][j] <= a[i][j-1] && a[i][j] <= a[i-1][j-1] && a[i][j] <= a[i+1][j+1] && a[i][j] <= a[i-1][j+1] && a[i][j] <= a[i+1][j-1])
            {
                cnt++;
            }
        }
    }
    printf("%d",cnt);
    return 0;
}

这个就是细节特判边界条件，代码量还不是很多，还可以0ms，嗯嗯嗯，不错不错。
只要吧边界的值弄成极限便可AC