预处理没点向上最大伸展的1 num[i][j]

然后对于每行，按num[i][j]从大到小排序，即以他们为底最大1矩阵

1 #include
      
        2 #include
       
         3 #include
        
          4 using namespace std; 5 int a[1005][1005],num[1005][1005]; 6 int cmp(int x,int y) 7 { 8   return x>y; 9 }10 int main()11 {12   int n,m,i,j,maxx;13   while (~scanf("%d%d",&n,&m))14   {15     memset(num,0,sizeof(num));16     for (i=1;i<=n;i++)17       for (j=1;j<=m;j++)18       {19         scanf("%1d",&a[i][j]);20         if (a[i][j]) num[i][j]=num[i-1][j]+1; 21       }22     maxx=0;23     for (i=1;i<=n;i++)24     {25       sort(num[i]+1,num[i]+m+1,cmp);26       for (j=1;j<=m;j++)27         if (num[i][j]*j>maxx) maxx=num[i][j]*j;28     }29     printf("%d\n",maxx);30   }31 }
        
       
      

题目链接：